দশম শ্রেণী – নবম অধ্যায় : দ্বিঘাত করনী সম্পূর্ণ সমাধান

কষে দেখি – 9.1

1.  মূলদ ও অমূলদ সংখ্যার গুণফল আকারে লিখি –

\[\left( i \right)\,\sqrt{175}\,\,\,\,\left( ii \right)\,2\sqrt{112}\,\,\,\,\left( iii \right)\,\sqrt{108}\,\,\,\,\left( iv \right)\,\sqrt{125}\,\,\,\,\left( v \right)\,5\sqrt{119}\]

উত্তর –

(i) \sqrt{175}\,=\sqrt{5\times 5\times 7}=5\sqrt{7}

এক্ষেত্রে 5 হল মূলদ সংখ্যা এবং\sqrt{7} হল অমূলদ সংখ্যা।

(ii) 2\sqrt{112}\,=2\sqrt{2\times 2\times 2\times 2\times 7}=2\times 2\times 2\sqrt{7}=8\sqrt{7}

এক্ষেত্রে 8 হল মূলদ সংখ্যা এবং \sqrt{7} হল অমূলদ সংখ্যা।

(iii) \sqrt{108}\,=\,\sqrt{2\times 2\times 3\times 3\times 3}=2\times 3\sqrt{3}=6\sqrt{3}

এক্ষেত্রে 6 হল মূলদ সংখ্যা এবং \sqrt{3} হল অমূলদ সংখ্যা।

(iv) \sqrt{125}\,=\sqrt{5\times 5\times 5}=5\sqrt{5}

এক্ষেত্রে 5 হল মূলদ সংখ্যা এবং \sqrt{5} হল অমূলদ সংখ্যা।

(v) 5\sqrt{119}=5\sqrt{119}

এক্ষেত্রে 5 হল মূলদ সংখ্যা এবং \sqrt{119} হল অমূলদ সংখ্যা।

2. প্রমাণ করি যে, \sqrt{108}-\sqrt{75}\,=\sqrt{3}

উত্তর –

বামপক্ষ, \sqrt{108}-\sqrt{75}\,
= \sqrt{2\times 2\times 3\times 3\times 3}-\sqrt{5\times 5\times 3}
= 2\times 3\sqrt{3}-5\sqrt{3}
= 6\sqrt{3}-5\sqrt{3}
= \sqrt{3} = ডানপক্ষ
∴ বামপক্ষ = ডানপক্ষ [প্রমাণিত]

3. দেখাই যে, \sqrt{98}+\sqrt{8}-2\sqrt{32}=\sqrt{2}

উত্তর –

বামপক্ষ, \sqrt{98}+\sqrt{8}-2\sqrt{32}
= \sqrt{2\times 7\times 7}+\sqrt{2\times 2\times 2}-2\sqrt{2\times 2\times 2\times 2\times 2}
= 7\sqrt{2}+2\sqrt{2}-2\times 2\times 2\sqrt{2}
= 9\sqrt{2}-8\sqrt{2}
= \sqrt{2} = ডানপক্ষ
∴ বামপক্ষ = ডানপক্ষ [প্রমাণিত]

4. দেখাই যে, 3\sqrt{48}-4\sqrt{75}+\sqrt{192}=0

উত্তর –

বামপক্ষ, 3\sqrt{48}-4\sqrt{75}+\sqrt{192}
= 3\sqrt{2\times 2\times 2\times 2\times 3}-4\sqrt{3\times 5\times 5}+\sqrt{3\times 2\times 2\times 2\times 2\times 2\times 2}
= 3\times 2\times 2\sqrt{3}-4\times 5\sqrt{3}+2\times 2\times 2\sqrt{3}
= 12\sqrt{3}-20\sqrt{3}+8\sqrt{3}
= 20\sqrt{3}-20\sqrt{3}
= 0 = ডানপক্ষ
∴ বামপক্ষ = ডানপক্ষ [প্রমাণিত]

5. সরলতম মান নির্নয় করি – \sqrt{12}+\sqrt{18}+\sqrt{27}-\sqrt{32}

উত্তর –  \sqrt{12}+\sqrt{18}+\sqrt{27}-\sqrt{32}
= \sqrt{3\times 4}+\sqrt{2\times 9}+\sqrt{3\times 9}-\sqrt{2\times 16}
= 2\sqrt{3}+3\sqrt{2}+3\sqrt{3}-4\sqrt{2}
= 2\sqrt{3}+3\sqrt{3}+3\sqrt{2}-4\sqrt{2}
= 5\sqrt{3}-\sqrt{2}
∴ নির্নেয় সরলতম মান – 5\sqrt{3}-\sqrt{2}

6. (a) \sqrt{5}+\sqrt{3} -এর সঙ্গে কত যোগ করলে যোগফল 2\sqrt{5} হবে, হিসাব করে লিখি।

উত্তর –  

মনেকরি p যোগ করতে হবে,
শর্তানুসারে, \sqrt{5}+\sqrt{3}+p=2\sqrt{5}
বা, p=2\sqrt{5}-\sqrt{5}-\sqrt{3}
∴ p=\sqrt{5}-\sqrt{3}
∴ \left( \sqrt{5}-\sqrt{3} \right) যোগ করতে হবে।

(b) 7-\sqrt{3} -এর থেকে কত বিয়োগ করলে বিয়োগফল 3+\sqrt{3} হবে, নির্নয় করি।

উত্তর –  

মনেকরি p বিয়োগ করতে হবে,
শর্তানুসারে, 7-\sqrt{3}-p=3+\sqrt{3}
বা, -p=3+\sqrt{3}-7+\sqrt{3}
বা, -p=-4+2\sqrt{3}
বা, -p=-\left( 4-2\sqrt{3} \right)
∴ p=4-2\sqrt{3}
∴ \left( 4-2\sqrt{3} \right) বিয়োগ করতে হবে।

(c) 2+\sqrt{3},\,\sqrt{3}+\sqrt{5} এবং 2+\sqrt{7} -এর যোগফল লিখি।

উত্তর –

2+\sqrt{3}+\,\sqrt{3}+\sqrt{5}+2+\sqrt{7}=2+2+\sqrt{3}+\,\sqrt{3}+\sqrt{5}+\sqrt{7}=4+2\sqrt{3}+\sqrt{5}+\sqrt{7}
∴ নির্নেয় যোগফল = 4+2\sqrt{3}+\sqrt{5}+\sqrt{7}

(d) \left( 10-\sqrt{11} \right) থেকে \left( -5+3\sqrt{11} \right) বিয়োগ করি ও বিয়োগফল লিখি।

উত্তর –

\left( 10-\sqrt{11} \right)-\left( -5+3\sqrt{11} \right)=10-\sqrt{11}+5-3\sqrt{11}=15-4\sqrt{11}
∴ নির্নেয় বিয়োগফল = \left( 15-4\sqrt{11} \right)

(e) \left( -5+\sqrt{7} \right)এবং \left( \sqrt{7}+\sqrt{2} \right) -এর যোগফল থেকে \left( 5+\sqrt{2}+\sqrt{7} \right) বিয়োগ করে বিয়োগফল নির্নয় করি।

উত্তর –

\[\left\{ \left( -5+\sqrt{7} \right)+\left( \sqrt{7}+\sqrt{2} \right) \right\}-\left( 5+\sqrt{2}+\sqrt{7} \right)\]

\[=\left\{ -5+\sqrt{7}+\sqrt{7}+\sqrt{2} \right\}-5-\sqrt{2}-\sqrt{7}\]

\[=-5+2\sqrt{7}+\sqrt{2}-5-\sqrt{2}-\sqrt{7}\]

\[=-10+\sqrt{7}\]

\[=\sqrt{7}-10\]

∴ নির্নেয় বিয়োগফল = \left( \sqrt{7}-10 \right)

(f) দুটি দ্বিঘাত করণী লিখি যাদের সমষ্টি মূলদ সংখ্যা।

উত্তর –

মনেকরি, দুটি দ্বিঘাত করণী যথাক্রমে \left( 2+\sqrt{3} \right) এবং \left( 2-\sqrt{3} \right)
তাদের সমষ্টি = \left( 2+\sqrt{3} \right)+\left( 2-\sqrt{3} \right)=2+\sqrt{3}+2-\sqrt{3}=4
সুতরাং, দ্বিঘাত করণী দুটির সমষ্টি মূলদ সংখ্যা।

কষে দেখি – 9.2

1. (a) {{3}^{\frac{1}{2}}}ও \sqrt{3} -এর গুণফল নির্নয় করি।

উত্তর –

\[{{3}^{\frac{1}{2}}}\times \sqrt{3}=\sqrt{3}\times \sqrt{3}=3\]

∴ নির্নেয় গুণফল = 3।

(b) 2\sqrt{2} -কে কত দিয়ে গুন করলে 4 পাব লিখি।

উত্তর –

মনেকরি x দিয়ে গুন করতে হবে।

শর্তানুসারে,

\[2\sqrt{2}\times x=4\]

\[\Rightarrow x=\frac{4}{2\sqrt{2}}=\frac{2}{\sqrt{2}}\]

\[\Rightarrow x=\frac{2\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}\]

∴ 2\sqrt{2} -কে \sqrt{2} দিয়ে গুন করলে 4 পাব।

(c) 3\sqrt{5} এবং 5\sqrt{3} -এর গুনফল নির্নয় করি।

উত্তর –

\[3\sqrt{5}\times 5\sqrt{3}=3\times 5\sqrt{5\times 3}=15\sqrt{15}\]

∴ নির্নেয় গুণফল = 15\sqrt{15}

(d) \sqrt{6}\times \sqrt{15}=x\sqrt{10} হলে, x -এর মান হিসাব করে লিখি।

উত্তর –

\[\sqrt{6}\times \sqrt{15}=x\sqrt{10}\]

\[\Rightarrow x=\frac{\sqrt{6}\times \sqrt{15}}{\sqrt{10}}\]

\[\Rightarrow x=\frac{\sqrt{2}\times \sqrt{3}\times \sqrt{3}\times \sqrt{5}}{\sqrt{2}\times \sqrt{5}}\]

\[\therefore \,\,x=3\]

∴ নির্নেয় x -এর মান = 3।

(e) \left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)=25-{{x}^{2}}একটি সমীকরণ হলে, x -এর মান হিসাব করে লিখি।

উত্তর –

\[\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)=25-{{x}^{2}}\]

\[\Rightarrow {{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=25-{{x}^{2}}\]

\[\Rightarrow 5-3=25-{{x}^{2}}\]

\[\Rightarrow {{x}^{2}}=25-5+3=23\]

\[\therefore \,\,x=\pm \sqrt{23}\]

∴ নির্নেয় x -এর মান = \pm \sqrt{23}

2. গুণফল নির্নয় করি –

(a) \sqrt{7}\times \sqrt{14}

উত্তর –

\[\sqrt{7}\times \sqrt{14}=\sqrt{7\times 7\times 2}=7\sqrt{2}\]

∴ নির্নেয় গুণফল = 7\sqrt{2}

(b) \sqrt{12}\times 2\sqrt{3}

উত্তর –

\[\sqrt{12}\times 2\sqrt{3}=\sqrt{2\times 2\times 3}\times 2\sqrt{3}=2\sqrt{3}\times 2\sqrt{3}=4\times 3=12\]

∴ নির্নেয় গুণফল = 12।

(c) \sqrt{5}\times \sqrt{15}\times \sqrt{3}

উত্তর –

\[\sqrt{5}\times \sqrt{15}\times \sqrt{3}=\sqrt{5\times 5\times 3\times 3}=5\times 3=15\]

∴ নির্নেয় গুণফল = 15।

(d) \sqrt{2}\left( 3+\sqrt{5} \right)

উত্তর –

\[\sqrt{2}\left( 3+\sqrt{5} \right)=3\sqrt{2}+\sqrt{10}\]

∴ নির্নেয় গুণফল = 3\sqrt{2}+\sqrt{10}

(e) \left( \sqrt{2}+\sqrt{3} \right)\left( \sqrt{2}-\sqrt{3} \right)

উত্তর –

\[\left( \sqrt{2}+\sqrt{3} \right)\left( \sqrt{2}-\sqrt{3} \right)={{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=2-3=-1\]

∴ নির্নেয় গুণফল = – 1।

(f) \left( 2\sqrt{3}+3\sqrt{2} \right)\left( 4\sqrt{2}+\sqrt{5} \right)

উত্তর –

\[\left( 2\sqrt{3}+3\sqrt{2} \right)\left( 4\sqrt{2}+\sqrt{5} \right)\]

\[=8\sqrt{6}+2\sqrt{15}+12\times 2+3\sqrt{10}\]

\[=24+8\sqrt{6}+2\sqrt{15}+3\sqrt{10}\]

∴ নির্নেয় গুণফল = 24+8\sqrt{6}+2\sqrt{15}+3\sqrt{10}

(g) \left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)\left( 2-\sqrt{3} \right)\left( 4+2\sqrt{3} \right)

উত্তর –

\[\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)\left( 2-\sqrt{3} \right)\left( 4+2\sqrt{3} \right)\]

\[=\left\{ {{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}} \right\}\left( 2-\sqrt{3} \right)\times 2\left( 2+\sqrt{3} \right)\]

\[=\left\{ 3-1 \right\}\times 2\left\{ {{\left( 2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right\}\]

\[=2\times 2\left\{ 4-3 \right\}=2\times 2\times 1=4\]

∴ নির্নেয় গুণফল = 4।

3. (a) \sqrt{5} -এর করনী নিরসক উৎপাদক \sqrt{x} হলে, x –এর ক্ষুদ্রতম মান কত হবে তা হিসাব করে লিখি। [যেখানে x একটি পূর্নসংখ্যা]

উত্তর –

মনেকরি, \sqrt{5} -এর করনী নিরসক উৎপাদক k\sqrt{5} [ যেখানে, k হল অশূন্য মূলদ সংখ্যা ]

সুতরাং, \sqrt{x}=k\sqrt{5}

এক্ষেত্রে, x –এর ক্ষুদ্রতম মান হবে যদি k = 1 হয়।

বা, \sqrt{x}=\sqrt{5}

∴ x = 5

∴ x –এর ক্ষুদ্রতম মান 5 হবে।

(b) 3\sqrt{2}\div 3 -এর মান নির্নয় করি।

উত্তর –

\[3\sqrt{2}\div 3=\frac{3\sqrt{2}}{3}=\sqrt{2}\]

∴ নির্নেয় মান = \sqrt{2}

(c) 7\div \sqrt{48}-এর হরের করনী নিরসন করতে হরকে ন্যূনতম কত দিয়ে গুন করতে হবে তা লিখি।

উত্তর –

\[7\div \sqrt{48}=\frac{7}{\sqrt{48}}=\frac{7}{\sqrt{3\times 16}}=\frac{7}{4\sqrt{3}}=\frac{7\times \sqrt{3}}{4\sqrt{3}\times \sqrt{3}}=\frac{7\sqrt{3}}{12}\]

∴ হরকে \sqrt{3} দিয়ে গুন করতে হবে।

(d) \left( \sqrt{5}+2 \right) -এর করনী নিরসক উৎপাদক নির্নয় করি যা করণীটির অনুবন্ধী করণী।

উত্তর –

\left( \sqrt{5}+2 \right) -এর করনী নিরসক উৎপাদক \left( -\sqrt{5}+2 \right) যা করণীটির অনুবন্ধী করণী।

(e) \left( \sqrt{5}+\sqrt{2} \right)\div \sqrt{7}=\frac{1}{7}\left( \sqrt{35}+a \right) হলে, a –এর মান নির্নয় করি।

উত্তর –

\[\left( \sqrt{5}+\sqrt{2} \right)\div \sqrt{7}=\frac{1}{7}\left( \sqrt{35}+a \right)\]

\[\Rightarrow \frac{\left( \sqrt{5}+\sqrt{2} \right)}{\sqrt{7}}=\frac{1}{7}\left( \sqrt{35}+a \right)\]

\[\Rightarrow \frac{\left( \sqrt{5}+\sqrt{2} \right)\times \sqrt{7}}{\sqrt{7}\times \sqrt{7}}=\frac{1}{7}\left( \sqrt{35}+a \right)\]

\[\Rightarrow \frac{\sqrt{35}+\sqrt{14}}{7}=\frac{\sqrt{35}+a}{7}\]

\[\therefore \,\,a=\sqrt{14}\]

∴ নির্নেয় a –এর মান = \sqrt{14}

(f) \frac{5}{\sqrt{3}-2} -এর হরের একটি করণী নিরসক উৎপাদক লিখি যা অনুবন্ধী করণী নয়।

উত্তর –

\[\frac{5}{\sqrt{3}-2}=\frac{5\left( \sqrt{3}+2 \right)}{\left( \sqrt{3}-2 \right)\left( \sqrt{3}+2 \right)}=\frac{5\left( \sqrt{3}+2 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 2 \right)}^{2}}}=\frac{5\left( \sqrt{3}+2 \right)}{3-4}=-5\left( \sqrt{3}+2 \right)\]

∴ \frac{5}{\sqrt{3}-2} -এর হরের একটি করণী নিরসক উৎপাদক হল \left( \sqrt{3}+2 \right)যা অনুবন্ধী করণী নয়।

4. \left( 9-4\sqrt{5} \right) ও \left( -2-\sqrt{7} \right) মিশ্র দ্বিঘাত করণীদ্বয়ের অনুবন্ধী করণীদ্বয় লিখি।

উত্তর –

\left( 9-4\sqrt{5} \right)-এর অনুবন্ধী করণী \left( 9+4\sqrt{5} \right)

\left( -2-\sqrt{7} \right)-এর অনুবন্ধী করণী \left( -2+\sqrt{7} \right)

5. নীচের মিশ্র দ্বিঘাত করণীর 2 টি করে করণী নিরসক উৎপাদক লিখি –

(i) \sqrt{5}+\sqrt{2}

উত্তর –

\sqrt{5}+\sqrt{2} -এর 2 টি করে করণী নিরসক উৎপাদক যথাক্রমে \left( \sqrt{5}-\sqrt{2} \right) এবং \left( -\sqrt{5}+\sqrt{2} \right)।

(ii) 13+\sqrt{6}

উত্তর –

13+\sqrt{6} -এর 2 টি করে করণী নিরসক উৎপাদক যথাক্রমে \left( 13-\sqrt{6} \right) এবং \left( -13+\sqrt{6} \right)।

(iii) \sqrt{8}-3

উত্তর –

\sqrt{8}-3 -এর 2 টি করে করণী নিরসক উৎপাদক যথাক্রমে \left( -\sqrt{8}-3 \right) এবং \left( \sqrt{8}+3 \right)।

(iv) \sqrt{17}-\sqrt{15}

উত্তর –

\sqrt{17}-\sqrt{15} -এর 2 টি করে করণী নিরসক উৎপাদক যথাক্রমে \left( \sqrt{17}+\sqrt{15} \right) এবং \left( -\sqrt{17}-\sqrt{15} \right)।

6. হরের করণী নিরসন করি –

(i) \frac{2\sqrt{3}+3\sqrt{2}}{\sqrt{6}}

উত্তর –

\[\frac{2\sqrt{3}+3\sqrt{2}}{\sqrt{6}}=\frac{\left( 2\sqrt{3}+3\sqrt{2} \right)\times \sqrt{6}}{\sqrt{6}\times \sqrt{6}}=\frac{2\sqrt{18}+3\sqrt{12}}{6}=\frac{6\sqrt{2}+6\sqrt{3}}{6}=\frac{6\left( \sqrt{2}+\sqrt{3} \right)}{6}=\sqrt{2}+\sqrt{3}\]

(ii) \frac{\sqrt{2}-1+\sqrt{6}}{\sqrt{5}}

উত্তর –

\[\frac{\sqrt{2}-1+\sqrt{6}}{\sqrt{5}}=\frac{\left( \sqrt{2}-1+\sqrt{6} \right)\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{\sqrt{10}-\sqrt{5}+\sqrt{30}}{5}\]

(iii) \frac{\sqrt{3}+1}{\sqrt{3}-1}

উত্তর –

\[\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left( \sqrt{3}+1 \right)\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}=\frac{3+\sqrt{3}+\sqrt{3}+1}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\frac{4+2\sqrt{3}}{3-1}=\frac{2\left( 2+\sqrt{3} \right)}{2}=2+\sqrt{3}\]

(iv) \frac{3+\sqrt{5}}{\sqrt{7}-\sqrt{3}}

উত্তর –

\[\frac{3+\sqrt{5}}{\sqrt{7}-\sqrt{3}}=\frac{\left( 3+\sqrt{5} \right)\left( \sqrt{7}+\sqrt{3} \right)}{\left( \sqrt{7}-\sqrt{3} \right)\left( \sqrt{7}+\sqrt{3} \right)}=\frac{3\sqrt{7}+3\sqrt{3}+\sqrt{35}+\sqrt{15}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}\]

\[=\frac{3\sqrt{7}+3\sqrt{3}+\sqrt{35}+\sqrt{15}}{7-3}=\frac{3\sqrt{7}+3\sqrt{3}+\sqrt{35}+\sqrt{15}}{4}\]

(v) \frac{3\sqrt{2}+1}{2\sqrt{5}-1}

উত্তর –

\[\frac{3\sqrt{2}+1}{2\sqrt{5}-1}=\frac{\left( 3\sqrt{2}+1 \right)\left( 2\sqrt{5}+1 \right)}{\left( 2\sqrt{5}-1 \right)\left( 2\sqrt{5}+1 \right)}=\frac{6\sqrt{10}+3\sqrt{2}+2\sqrt{5}+1}{{{\left( 2\sqrt{5} \right)}^{2}}-{{\left( 1 \right)}^{2}}}\]

\[=\frac{6\sqrt{10}+3\sqrt{2}+2\sqrt{5}+1}{20-1}=\frac{6\sqrt{10}+3\sqrt{2}+2\sqrt{5}+1}{19}\]

(vi) \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}

উত্তর –

\[\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=\frac{\left( 3\sqrt{2}+2\sqrt{3} \right)\left( 3\sqrt{2}+2\sqrt{3} \right)}{\left( 3\sqrt{2}-2\sqrt{3} \right)\left( 3\sqrt{2}+2\sqrt{3} \right)}=\frac{18+6\sqrt{6}+6\sqrt{6}+12}{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}}}\]

\[=\frac{30+12\sqrt{6}}{18-12}=\frac{6\left( 5+2\sqrt{6} \right)}{6}=5+2\sqrt{6}\]

7. প্রথমটিকে দ্বিতীয়টি দিয়ে ভাগ করে ভাজককে মূলদ সংখ্যায় পরিণত করি।

(i) 3\sqrt{2}+\sqrt{5},\,\sqrt{2}+1

উত্তর –

\[\frac{3\sqrt{2}+\sqrt{5}}{\sqrt{2}+1}=\frac{\left( 3\sqrt{2}+\sqrt{5} \right)\left( \sqrt{2}-1 \right)}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}=\frac{6-3\sqrt{2}+\sqrt{10}-\sqrt{5}}{2-1}=6-3\sqrt{2}+\sqrt{10}-\sqrt{5}\]

(ii) 2\sqrt{3}-\sqrt{2},\,\sqrt{2}-\sqrt{3}

উত্তর –

\[\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}=\frac{\left( 2\sqrt{3}-\sqrt{2} \right)\left( \sqrt{2}+\sqrt{3} \right)}{\left( \sqrt{2}-\sqrt{3} \right)\left( \sqrt{2}+\sqrt{3} \right)}=\frac{2\sqrt{6}+6-2-\sqrt{6}}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}=\frac{4+\sqrt{6}}{2-3}=-\left( 4+\sqrt{6} \right)\]

(iii) 3+\sqrt{6},\,\sqrt{3}+\sqrt{2}

উত্তর –

\[\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{\left( 3+\sqrt{6} \right)\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}=\frac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12}}{3-2}\]

\[=\frac{3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}{1}=\sqrt{3}\]

8. মান নির্নয় করি –

(i) \frac{2\sqrt{5}+1}{\sqrt{5}+1}-\frac{4\sqrt{5}-1}{\sqrt{5}-1}

উত্তর –

\[\frac{2\sqrt{5}+1}{\sqrt{5}+1}-\frac{4\sqrt{5}-1}{\sqrt{5}-1}\]

\[=\frac{\left( 2\sqrt{5}+1 \right)\left( \sqrt{5}-1 \right)}{\left( \sqrt{5}+1 \right)\left( \sqrt{5}-1 \right)}-\frac{\left( 4\sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}\]

\[=\frac{10-2\sqrt{5}+\sqrt{5}-1}{5-1}-\frac{20+4\sqrt{5}-\sqrt{5}-1}{5-1}\]

\[=\frac{9-\sqrt{5}}{4}-\frac{19+3\sqrt{5}}{4}\]

\[=\frac{9-\sqrt{5}-19-3\sqrt{5}}{4}\]

\[=\frac{-10-4\sqrt{5}}{4}=\frac{-2\left( 5+2\sqrt{5} \right)}{4}=\frac{-\left( 5+2\sqrt{5} \right)}{2}\]

(ii) \frac{8+3\sqrt{2}}{3+\sqrt{5}}-\frac{8-3\sqrt{2}}{3-\sqrt{5}}

উত্তর –

\[\frac{8+3\sqrt{2}}{3+\sqrt{5}}-\frac{8-3\sqrt{2}}{3-\sqrt{5}}\]

\[=\frac{\left( 8+3\sqrt{2} \right)\left( 3-\sqrt{5} \right)}{\left( 3+\sqrt{5} \right)\left( 3-\sqrt{5} \right)}-\frac{\left( 8-3\sqrt{2} \right)\left( 3+\sqrt{5} \right)}{\left( 3-\sqrt{5} \right)\left( 3+\sqrt{5} \right)}\]

\[=\frac{24-8\sqrt{5}+9\sqrt{2}-3\sqrt{10}}{9-5}-\frac{24+8\sqrt{5}-9\sqrt{2}-3\sqrt{10}}{9-5}\]

\[=\frac{24-8\sqrt{5}+9\sqrt{2}-3\sqrt{10}}{4}-\frac{24+8\sqrt{5}-9\sqrt{2}-3\sqrt{10}}{4}\]

\[=\frac{24-8\sqrt{5}+9\sqrt{2}-3\sqrt{10}-24-8\sqrt{5}+9\sqrt{2}+3\sqrt{10}}{4}\]

\[=\frac{18\sqrt{2}-16\sqrt{5}}{4}=\frac{2\left( 9\sqrt{2}-8\sqrt{5} \right)}{4}=\frac{\left( 9\sqrt{2}-8\sqrt{5} \right)}{2}\]

কষে দেখি – 9.3

1. (a) m+\frac{1}{m}=\sqrt{3} হলে \left( i \right)\,\,{{m}^{2}}+\frac{1}{{{m}^{2}}} এবং \left( ii \right)\,\,{{m}^{3}}+\frac{1}{{{m}^{3}}}-এদের সরলতম মান নির্নয় করি।

উত্তর –

দেওয়া আছে যে, m+\frac{1}{m}=\sqrt{3}

\[\left( i \right)\,\,{{m}^{2}}+\frac{1}{{{m}^{2}}}={{\left( m+\frac{1}{m} \right)}^{2}}-2.m.\frac{1}{m}={{\left( \sqrt{3} \right)}^{2}}-2=3-2=1\]

\[\left( ii \right)\,\,{{m}^{3}}+\frac{1}{{{m}^{3}}}={{\left( m+\frac{1}{m} \right)}^{3}}-3.m.\frac{1}{m}\left( m+\frac{1}{m} \right)={{\left( \sqrt{3} \right)}^{3}}-3\left( \sqrt{3} \right)=3\sqrt{3}-3\sqrt{3}=0\]

(b) দেখাই যে, \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}

উত্তর –

বামপক্ষ = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}

= \frac{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}+\sqrt{3} \right)}{\left( \sqrt{5}-\sqrt{3} \right)\left( \sqrt{5}+\sqrt{3} \right)}-\frac{\left( \sqrt{5}-\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)}

= \frac{{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}-\frac{{{\left( \sqrt{5}-\sqrt{3} \right)}^{2}}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}

= \frac{5+2\sqrt{15}+3}{5-3}-\frac{5-2\sqrt{15}+3}{5-3}

= \frac{8+2\sqrt{15}}{2}-\frac{8-2\sqrt{15}}{2}

= \frac{8+2\sqrt{15}-8+2\sqrt{15}}{2}

= \frac{4\sqrt{15}}{2}=2\sqrt{15} = ডানপক্ষ [প্রমাণিত]

2. সরল করি –

(a) \frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}

উত্তর –

\[\frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}\]

\[=\frac{\sqrt{2}\left( 2+\sqrt{3} \right)\times \sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)\times \sqrt{3}\left( \sqrt{3}-1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)\times \sqrt{3}\left( \sqrt{3}+1 \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)\times \sqrt{3}\left( \sqrt{3}+1 \right)}\]

\[=\frac{\sqrt{2}\times \sqrt{3}\left( 2+\sqrt{3} \right)\left( \sqrt{3}-1 \right)}{\sqrt{3}\times \sqrt{3}\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}-\frac{\sqrt{2}\times \sqrt{3}\left( 2-\sqrt{3} \right)\left( \sqrt{3}+1 \right)}{\sqrt{3}\times \sqrt{3}\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}\]

\[=\frac{\sqrt{6}\left( 2\sqrt{3}-2+3-\sqrt{3} \right)}{3\left\{ {{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}} \right\}}-\frac{\sqrt{6}\left( 2\sqrt{3}+2-3-\sqrt{3} \right)}{3\left\{ {{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}} \right\}}\]

\[=\frac{\sqrt{6}\left( \sqrt{3}+1 \right)}{3\times 2}-\frac{\sqrt{6}\left( \sqrt{3}-1 \right)}{3\times 2}\]

\[=\frac{\sqrt{18}+\sqrt{6}}{6}-\frac{\sqrt{18}-\sqrt{6}}{6}\]

\[=\frac{\sqrt{18}+\sqrt{6}-\sqrt{18}+\sqrt{6}}{6}\]

\[=\frac{2\sqrt{6}}{6}=\frac{\sqrt{6}}{3}\]

∴ নির্নেয় সরলফল = \frac{\sqrt{6}}{3}

(b) \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}

উত্তর –

\[\frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}\]

\[=\frac{3\sqrt{7}\left( \sqrt{5}-\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)}-\frac{5\sqrt{5}\left( \sqrt{2}-\sqrt{7} \right)}{\left( \sqrt{2}+\sqrt{7} \right)\left( \sqrt{2}-\sqrt{7} \right)}+\frac{2\sqrt{2}\left( \sqrt{7}-\sqrt{5} \right)}{\left( \sqrt{7}+\sqrt{5} \right)\left( \sqrt{7}-\sqrt{5} \right)}\]

\[=\frac{3\left( \sqrt{35}-\sqrt{14} \right)}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}-\frac{5\left( \sqrt{10}-\sqrt{35} \right)}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{7} \right)}^{2}}}+\frac{2\left( \sqrt{14}-\sqrt{10} \right)}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}\]

\[=\frac{3\left( \sqrt{35}-\sqrt{14} \right)}{5-2}-\frac{5\left( \sqrt{10}-\sqrt{35} \right)}{2-7}+\frac{2\left( \sqrt{14}-\sqrt{10} \right)}{7-5}\]

\[=\frac{3\left( \sqrt{35}-\sqrt{14} \right)}{3}-\frac{5\left( \sqrt{10}-\sqrt{35} \right)}{-5}+\frac{2\left( \sqrt{14}-\sqrt{10} \right)}{2}\]

\[=\left( \sqrt{35}-\sqrt{14} \right)+\left( \sqrt{10}-\sqrt{35} \right)+\left( \sqrt{14}-\sqrt{10} \right)\]

\[=\sqrt{35}-\sqrt{14}+\sqrt{10}-\sqrt{35}+\sqrt{14}-\sqrt{10}\,\,=\,\,0\]

∴ নির্নেয় সরলফল = 0

(c) \frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}

উত্তর –

\[\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}\]

\[=\frac{4\sqrt{3}\left( 2+\sqrt{2} \right)}{\left( 2-\sqrt{2} \right)\left( 2+\sqrt{2} \right)}-\frac{30\left( 4\sqrt{3}+\sqrt{18} \right)}{\left( 4\sqrt{3}-\sqrt{18} \right)\left( 4\sqrt{3}+\sqrt{18} \right)}-\frac{\sqrt{18}\left( 3+\sqrt{12} \right)}{\left( 3-\sqrt{12} \right)\left( 3+\sqrt{12} \right)}\]

\[=\frac{4\left( 2\sqrt{3}+\sqrt{6} \right)}{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}-\frac{30\left( 4\sqrt{3}+3\sqrt{2} \right)}{{{\left( 4\sqrt{3} \right)}^{2}}-{{\left( \sqrt{18} \right)}^{2}}}-\frac{3\sqrt{2}\left( 3+2\sqrt{3} \right)}{{{\left( 3 \right)}^{2}}-{{\left( \sqrt{12} \right)}^{2}}}\]

\[=\frac{4\left( 2\sqrt{3}+\sqrt{6} \right)}{4-2}-\frac{30\left( 4\sqrt{3}+3\sqrt{2} \right)}{48-18}-\frac{3\sqrt{2}\left( 3+2\sqrt{3} \right)}{9-12}\]

\[=\frac{4\left( 2\sqrt{3}+\sqrt{6} \right)}{2}-\frac{30\left( 4\sqrt{3}+3\sqrt{2} \right)}{30}-\frac{3\sqrt{2}\left( 3+2\sqrt{3} \right)}{-3}\]

\[=2\left( 2\sqrt{3}+\sqrt{6} \right)-\left( 4\sqrt{3}+3\sqrt{2} \right)+\sqrt{2}\left( 3+2\sqrt{3} \right)\]

\[=4\sqrt{3}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}\]

\[=4\sqrt{6}\]

∴ নির্নেয় সরলফল =4\sqrt{6}

(d) \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}

উত্তর –

\[\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\]

\[=\frac{3\sqrt{2}\left( \sqrt{3}-\sqrt{6} \right)}{\left( \sqrt{3}+\sqrt{6} \right)\left( \sqrt{3}-\sqrt{6} \right)}-\frac{4\sqrt{3}\left( \sqrt{6}-\sqrt{2} \right)}{\left( \sqrt{6}+\sqrt{2} \right)\left( \sqrt{6}-\sqrt{2} \right)}+\frac{\sqrt{6}\left( \sqrt{2}-\sqrt{3} \right)}{\left( \sqrt{2}+\sqrt{3} \right)\left( \sqrt{2}-\sqrt{3} \right)}\]

\[=\frac{3\left( \sqrt{6}-\sqrt{12} \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}-\frac{4\left( \sqrt{18}-\sqrt{6} \right)}{{{\left( \sqrt{6} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}+\frac{\sqrt{12}-\sqrt{18}}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}\]

\[=\frac{3\left( \sqrt{6}-2\sqrt{3} \right)}{3-6}-\frac{4\left( 3\sqrt{2}-\sqrt{6} \right)}{6-2}+\frac{2\sqrt{3}-3\sqrt{2}}{2-3}\]

\[=\frac{3\left( \sqrt{6}-2\sqrt{3} \right)}{-3}-\frac{4\left( 3\sqrt{2}-\sqrt{6} \right)}{4}+\frac{2\sqrt{3}-3\sqrt{2}}{-1}\]

\[=-\left( \sqrt{6}-2\sqrt{3} \right)-\left( 3\sqrt{2}-\sqrt{6} \right)-\left( 2\sqrt{3}-3\sqrt{2} \right)\]

\[=-\sqrt{6}+2\sqrt{3}-3\sqrt{2}+\sqrt{6}-2\sqrt{3}+3\sqrt{2}\]

\[=\,0\]

∴ নির্নেয় সরলফল = 0

3. যদি x = 2, y = 3 এবং z = 6 হয় তবে, \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}} -এর মান হিসাব করে লিখি।

উত্তর –

\[\frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\]

\[=\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\,\,\left[ \because \,x=2,\,y=3,\,z=6 \right]\]

\[=\frac{3\sqrt{2}\left( \sqrt{3}-\sqrt{6} \right)}{\left( \sqrt{3}+\sqrt{6} \right)\left( \sqrt{3}-\sqrt{6} \right)}-\frac{4\sqrt{3}\left( \sqrt{6}-\sqrt{2} \right)}{\left( \sqrt{6}+\sqrt{2} \right)\left( \sqrt{6}-\sqrt{2} \right)}+\frac{\sqrt{6}\left( \sqrt{2}-\sqrt{3} \right)}{\left( \sqrt{2}+\sqrt{3} \right)\left( \sqrt{2}-\sqrt{3} \right)}\,\]

\[=\frac{3\left( \sqrt{6}-\sqrt{12} \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}-\frac{4\left( \sqrt{18}-\sqrt{6} \right)}{{{\left( \sqrt{6} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}+\frac{\sqrt{12}-\sqrt{18}}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}\]

\[=\frac{3\left( \sqrt{6}-2\sqrt{3} \right)}{3-6}-\frac{4\left( 3\sqrt{2}-\sqrt{6} \right)}{6-2}+\frac{2\sqrt{3}-3\sqrt{2}}{2-3}\]

\[=\frac{3\left( \sqrt{6}-2\sqrt{3} \right)}{-3}-\frac{4\left( 3\sqrt{2}-\sqrt{6} \right)}{4}+\frac{2\sqrt{3}-3\sqrt{2}}{-1}\]

\[=-\left( \sqrt{6}-2\sqrt{3} \right)-\left( 3\sqrt{2}-\sqrt{6} \right)-\left( 2\sqrt{3}-3\sqrt{2} \right)\]

\[=-\sqrt{6}+2\sqrt{3}-3\sqrt{2}+\sqrt{6}-2\sqrt{3}+3\sqrt{2}\]

\[=0\]

4. x=\sqrt{7}+\sqrt{6} হলে \left( i \right)\,\,x-\frac{1}{x}\,\,\left( ii \right)\,\,x+\frac{1}{x}\,\,\left( iii \right)\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}} এবং \left( iv \right)\,\,{{x}^{3}}+\frac{1}{{{x}^{3}}} –এদের সরলতম মান নির্নয় করি।

উত্তর –

দেওয়া আছে যে, x=\sqrt{7}+\sqrt{6}

\[\therefore \,\frac{1}{x}=\frac{1}{\sqrt{7}+\sqrt{6}}=\frac{\left( \sqrt{7}-\sqrt{6} \right)}{\left( \sqrt{7}+\sqrt{6} \right)\left( \sqrt{7}-\sqrt{6} \right)}=\frac{\left( \sqrt{7}-\sqrt{6} \right)}{7-6}=\sqrt{7}-\sqrt{6}\] 

\[\left( i \right)\,x-\frac{1}{x}\]

\[=\left( \sqrt{7}+\sqrt{6} \right)-\left( \sqrt{7}-\sqrt{6} \right)\]

\[=\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6}\]

\[=2\sqrt{6}\]

\[\left( ii \right)\,x+\frac{1}{x}\]

\[=\left( \sqrt{7}+\sqrt{6} \right)+\left( \sqrt{7}-\sqrt{6} \right)\]

\[=\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6}\]

\[=2\sqrt{7}\]

\[\left( iii \right)\,{{x}^{2}}+\frac{1}{{{x}^{2}}}\]

\[={{\left( x+\frac{1}{x} \right)}^{2}}-2.x.\frac{1}{x}\]

\[={{\left( 2\sqrt{7} \right)}^{2}}-2\]

\[=28-2=26\]

\[\left( iv \right)\,{{x}^{3}}+\frac{1}{{{x}^{3}}}\]

\[={{\left( x+\frac{1}{x} \right)}^{3}}-3.x.\frac{1}{x}\left( x+\frac{1}{x} \right)\]

\[={{\left( 2\sqrt{7} \right)}^{3}}-3\left( 2\sqrt{7} \right)\]

\[=56\sqrt{7}-6\sqrt{7}=50\sqrt{7}\]

5. সরল করি – \frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}} সরলফল 14 হলে, x –এর মান কী কী হবে হিসাব করে লিখি।

উত্তর –

\[\frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}\]

\[=\frac{\left( x+\sqrt{{{x}^{2}}-1} \right)\left( x+\sqrt{{{x}^{2}}-1} \right)}{\left( x-\sqrt{{{x}^{2}}-1} \right)\left( x+\sqrt{{{x}^{2}}-1} \right)}+\frac{\left( x-\sqrt{{{x}^{2}}-1} \right)\left( x-\sqrt{{{x}^{2}}-1} \right)}{\left( x+\sqrt{{{x}^{2}}-1} \right)\left( x-\sqrt{{{x}^{2}}-1} \right)}\]

\[=\frac{{{x}^{2}}+x\sqrt{{{x}^{2}}-1}+x\sqrt{{{x}^{2}}-1}+{{x}^{2}}-1}{{{\left( x \right)}^{2}}-{{\left( \sqrt{{{x}^{2}}-1} \right)}^{2}}}+\frac{{{x}^{2}}-x\sqrt{{{x}^{2}}-1}+-\sqrt{{{x}^{2}}-1}+{{x}^{2}}-1}{{{\left( x \right)}^{2}}-{{\left( \sqrt{{{x}^{2}}-1} \right)}^{2}}}\]

\[=\frac{2{{x}^{2}}-1+2x\sqrt{{{x}^{2}}-1}}{{{x}^{2}}-{{x}^{2}}+1}+\frac{2{{x}^{2}}-1-2x\sqrt{{{x}^{2}}-1}}{{{x}^{2}}-{{x}^{2}}+1}\]

\[=2{{x}^{2}}-1+2{{x}^{2}}-1\]

\[=4{{x}^{2}}-2\]

∴ নির্নেয় সরলফল =4{{x}^{2}}-2

প্রশ্নানুসারে,

\[4{{x}^{2}}-2=14\]

\[\Rightarrow 4{{x}^{2}}=14+2=16\]

\[\Rightarrow {{x}^{2}}=\frac{16}{4}=4\]

\[\therefore \,x=\pm \sqrt{4}=\pm 2\]

6. যদি ও হয়, তবে নীচের মান গুলি নির্নয় করি।

উত্তর –

\[\therefore \,\,\,a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{{{\left( \sqrt{5}+1 \right)}^{2}}+{{\left( \sqrt{5}-1 \right)}^{2}}}{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}\]

\[\Rightarrow a+b=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{5-1}=\frac{12}{4}=3\]

\[\therefore \,\,\,a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{{{\left( \sqrt{5}+1 \right)}^{2}}-{{\left( \sqrt{5}-1 \right)}^{2}}}{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}\]

\[\Rightarrow a-b=\frac{5+2\sqrt{5}+1-5+2\sqrt{5}-1}{5-1}=\frac{4\sqrt{5}}{4}=\sqrt{5}\]

\[\therefore \,\,\,ab=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1\]

(i) \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}

\[\left( i \right)\,\,\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}\]

\[=\frac{{{a}^{2}}+2ab+{{b}^{2}}-ab}{{{a}^{2}}-2ab+{{b}^{2}}+ab}\]

\[=\frac{{{\left( a+b \right)}^{2}}-ab}{{{\left( a-b \right)}^{2}}+ab}\]

\[=\frac{{{\left( 3 \right)}^{2}}-1}{{{\left( \sqrt{5} \right)}^{2}}+1}=\frac{9-1}{5+1}=\frac{8}{6}=\frac{4}{3}=1\frac{1}{3}\]

(ii) \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}

\[\left( ii \right)\,\,\frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}=\frac{{{\left( \sqrt{5} \right)}^{3}}}{{{\left( 3 \right)}^{3}}}=\frac{5\sqrt{5}}{27}\]

(iii) \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}

\[\left( iii \right)\,\,\frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}\]

\[=\frac{3\left( {{a}^{2}}+{{b}^{2}} \right)+5ab}{3\left( {{a}^{2}}+{{b}^{2}} \right)+5ab}\]

\[=\frac{3\left\{ {{\left( a+b \right)}^{2}}-2ab \right\}+5ab}{3\left\{ {{\left( a-b \right)}^{2}}+2ab \right\}-5ab}\]

\[=\frac{3\left\{ {{\left( 3 \right)}^{2}}-2\left( 1 \right) \right\}+5\left( 1 \right)}{3\left\{ {{\left( \sqrt{5} \right)}^{2}}+2\left( 1 \right) \right\}-5\left( 1 \right)}\]

\[=\frac{3\left\{ 9-2 \right\}+5}{3\left\{ 5+2 \right\}-5}\]

\[=\frac{3\left\{ 7 \right\}+5}{3\left\{ 7 \right\}-5}=\frac{21+5}{21-5}=\frac{26}{16}=\frac{13}{8}=1\frac{5}{8}\]

(iv) \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}

\[\left( iv \right)\,\,\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}\]

\[=\frac{{{\left( a+b \right)}^{3}}-3ab\left( a+b \right)}{{{\left( a-b \right)}^{3}}+3ab\left( a-b \right)}\]

\[=\frac{{{\left( 3 \right)}^{3}}-3\left( 1 \right)\left( 3 \right)}{{{\left( \sqrt{5} \right)}^{3}}+3\left( 1 \right)\left( \sqrt{5} \right)}\]

\[=\frac{27-9}{5\sqrt{5}+3\sqrt{5}}=\frac{18}{8\sqrt{5}}=\frac{9}{4\sqrt{5}}=\frac{9\times \sqrt{5}}{4\sqrt{5}\times \sqrt{5}}=\frac{9\sqrt{5}}{20}\]

7. যদি \[x=2+\sqrt{3},\,\,y=2-\sqrt{3}\] হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্নয় করি।

উত্তর –

দেওয়া আছে যে, x=2+\sqrt{3},\,\,y=2-\sqrt{3}

\[\therefore \,\,\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{\left( 2-\sqrt{3} \right)}{\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)}=\frac{\left( 2-\sqrt{3} \right)}{4-3}=2-\sqrt{3}\]

\[\therefore \,\,\frac{1}{y}=\frac{1}{2-\sqrt{3}}=\frac{\left( 2+\sqrt{3} \right)}{\left( 2-\sqrt{3} \right)\left( 2+\sqrt{3} \right)}=\frac{\left( 2+\sqrt{3} \right)}{4-3}=2+\sqrt{3}\]

\[\therefore \,\,\,xy=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=4-3=1\]

(a) (i) x-\frac{1}{x}

\[\left( a \right)\,\left( i \right)\,x-\frac{1}{x}=\left( 2+\sqrt{3} \right)-\left( 2-\sqrt{3} \right)=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\]

(ii) {{y}^{2}}+\frac{1}{{{y}^{2}}}

\[\left( ii \right)\,\,{{y}^{2}}+\frac{1}{{{y}^{2}}}={{\left( y+\frac{1}{y} \right)}^{2}}-2.y.\frac{1}{y}={{\left\{ 2-\sqrt{3}+2+\sqrt{3} \right\}}^{2}}-2={{\left\{ 4 \right\}}^{2}}-2=16-2=14\]

(iii) {{x}^{3}}-\frac{1}{{{x}^{3}}}

\[\left( iii \right)\,\,{{x}^{3}}-\frac{1}{{{x}^{3}}}={{\left( x-\frac{1}{x} \right)}^{3}}+3.x.\frac{1}{x}\left( x-\frac{1}{x} \right)={{\left( 2\sqrt{3} \right)}^{3}}+3\left( 2\sqrt{3} \right)=24\sqrt{3}+6\sqrt{3}=30\sqrt{3}\]

(iv) [xy+\frac{1}{xy}

\[\left( iv \right)\,\,xy+\frac{1}{xy}=1+\frac{1}{1}=1+1=2\]

\[\left( b \right)\,\,3{{x}^{2}}-5xy+3{{y}^{2}}\]

\[=3\left( {{x}^{2}}+{{y}^{2}} \right)-5xy\]

\[=3\left\{ {{\left( x+y \right)}^{2}}-2xy \right\}-5xy\]

\[=3\left\{ {{\left( 2+\sqrt{3}+2-\sqrt{3} \right)}^{2}}-2\left( 1 \right) \right\}-5\left( 1 \right)\]

\[=3\left\{ {{\left( 4 \right)}^{2}}-2 \right\}-5\]

\[=3\left\{ 16-2 \right\}-5=3\left\{ 14 \right\}-5=42-5=37\]

8. x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} এবং xy = 1 হলে, দেখাই যে, \frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}=\frac{12}{11}

উত্তর –

দেওয়া আছে যে, x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} এবং xy = 1

\[\therefore \,\,y=\frac{1}{x}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\]

\[\therefore \,\,x+y=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{{{\left( \sqrt{7}+\sqrt{3} \right)}^{2}}+{{\left( \sqrt{7}-\sqrt{3} \right)}^{2}}}{\left( \sqrt{7}-\sqrt{3} \right)\left( \sqrt{7}+\sqrt{3} \right)}\]

\[\Rightarrow x+y=\frac{7+2\sqrt{21}+3+7-2\sqrt{21}+3}{7-3}=\frac{20}{4}=5\]

\[\therefore \,\,x-y=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}-\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{{{\left( \sqrt{7}+\sqrt{3} \right)}^{2}}-{{\left( \sqrt{7}-\sqrt{3} \right)}^{2}}}{\left( \sqrt{7}-\sqrt{3} \right)\left( \sqrt{7}+\sqrt{3} \right)}\]

\[\Rightarrow x-y=\frac{7+2\sqrt{21}+3-7+2\sqrt{21}-3}{7-3}=\frac{4\sqrt{21}}{4}=\sqrt{21}\]

বামপক্ষ,

\[\frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}\]

\[=\frac{{{x}^{2}}+2xy+{{y}^{2}}-xy}{{{x}^{2}}-2xy+{{y}^{2}}+xy}\]

\[=\frac{{{\left( x+y \right)}^{2}}-xy}{{{\left( x-y \right)}^{2}}+xy}\]

\[=\frac{{{\left( 5 \right)}^{2}}-1}{{{\left( \sqrt{21} \right)}^{2}}+1}=\frac{25-1}{21+1}=\frac{24}{22}=\frac{12}{11}\]

∴ বামপক্ষ = ডানপক্ষ [ প্রমাণিত]

9. \left( \sqrt{7}+1 \right) এবং \left( \sqrt{5}+\sqrt{3} \right) -এর মধ্যে কোনটি বড়ো লিখি।

উত্তর –

\[{{\left( \sqrt{7}+1 \right)}^{2}}=7+2\sqrt{7}+1=8+2\sqrt{7}\]

\[{{\left( \sqrt{5}+\sqrt{3} \right)}^{2}}=5+2\sqrt{15}+3=8+2\sqrt{15}\]

যেহেতু, \sqrt{15}>\sqrt{7}, সুতরাং, 8+2\sqrt{15}>8+2\sqrt{7}

∴ \left( \sqrt{5}+\sqrt{3} \right) সংখ্যাটি বড়ো।

10. অতি সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q) –

(i) x=2+\sqrt{3} হলে, x+\frac{1}{x} -এর মান

(a) 2      (b)         (c) 4      (d) x=2-\sqrt{3}

উত্তর –

\[x=2+\sqrt{3}\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{\left( 2-\sqrt{3} \right)}{\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)}=\frac{\left( 2-\sqrt{3} \right)}{4-3}=2-\sqrt{3}\]

\[\therefore \,\,x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4\]

∴ নির্নেয় উত্তর হল – (c) 4

(ii) যদি p+q=\sqrt{13} এবং p-q=\sqrt{5}হয়, তাহলে pq –এর মান

(a) 2      (b) 18   (c) 9      (d) 8

উত্তর –

\[pq=\frac{{{\left( p+q \right)}^{2}}-{{\left( p-q \right)}^{2}}}{4}=\frac{{{\left( 13 \right)}^{2}}-{{\left( 5 \right)}^{2}}}{4}=\frac{8}{4}=2\]

∴ নির্নেয় উত্তর হল – (a) 2

(iii) যদি a+b=\sqrt{5} এবং a-b=\sqrt{3} হয়, তাহলে (a² + b²) –এর মান  

(a) 8      (b) 4     (c) 2      (d) 1

উত্তর –

\[{{a}^{2}}+{{b}^{2}}=\frac{{{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}}{2}=\frac{{{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}{2}=\frac{5+3}{2}=\frac{8}{2}=4\]

∴ নির্নেয় উত্তর হল – (b) 4

(iv) \sqrt{125} থেকে \sqrt{5} বিয়োগ করলে বিয়োগফল হবে

(a) \sqrt{80}    (b) \sqrt{120} (c) \sqrt{100}        (d) কোনটিই নয়

উত্তর –

\[\sqrt{125}-\sqrt{5}=5\sqrt{5}-\sqrt{5}=4\sqrt{5}=\sqrt{4\times 4\times 5}=\sqrt{80}\]

∴ নির্নেয় উত্তর হল – (a)

(v) \left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right) -এর গুণফল

(a) 22   (b) 44   (c) 2      (d) 11

উত্তর –

\[\left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right)\]

\[=\left\{ \left( 5-\sqrt{3} \right)\left( 5+\sqrt{3} \right) \right\}\left\{ \left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right) \right\}\]

\[=\left\{ 25-3 \right\}\left\{ 3-1 \right\}\]

\[=22\times 2=44\]

∴ নির্নেয় উত্তর হল – (b) 44

(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি –

(i) \sqrt{75} এবং \sqrt{147} সদৃশ করণী।

উত্তর –

\[\sqrt{75}=5\sqrt{3},\,\,\sqrt{147}=7\sqrt{3}\]

∴ সত্য।

(ii) [/katex][\sqrt{\pi }[/katex] একটি দ্বিঘাত করণী।

উত্তর – মিথ্যা

(C) শূন্যস্থান পূরণ করি –

(i) 5\sqrt{11} একটি __________ সংখ্যা। (মূলদ/ অমূলদ)

উত্তর – অমূলদ

(ii) \left( \sqrt{3}-5 \right) -এর অনুবন্ধী করণী __________।

উত্তর – \left( -\sqrt{3}-5 \right)

(iii) দুটি দ্বিঘাত করণীর যোগফল ও গুণফল একটি মূলদ সংখ্যা হলে করণীদ্বয় __________ করণী।

উত্তর – অনুবন্ধী করণী

11. সংক্ষিপ্ত উত্তর প্রশ্ন (S.A.)

(i) x=3+2\sqrt{2} হলে, x+\frac{1}{x} -এর মান লিখি।

উত্তর –

\[x=3+2\sqrt{2}\Rightarrow \frac{1}{x}=\frac{1}{3+2\sqrt{2}}=\frac{\left( 3-2\sqrt{2} \right)}{\left( 3+2\sqrt{2} \right)\left( 3-2\sqrt{2} \right)}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}\]

\[\therefore \,\,x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6\]

(ii) \left( \sqrt{15}+\sqrt{3} \right) এবং \left( \sqrt{10}+\sqrt{8} \right) -এর মধ্যে কোনটি বড়ো লিখি।

উত্তর –

\[{{\left( \sqrt{15}+\sqrt{3} \right)}^{2}}=15+2\sqrt{45}+3=18+6\sqrt{5}\]

\[{{\left( \sqrt{10}+\sqrt{8} \right)}^{2}}=10+2\sqrt{80}+8=18+8\sqrt{5}\]

\[\therefore \,\,8\sqrt{5}>6\sqrt{5}\Rightarrow 18+8\sqrt{5}>18+6\sqrt{5}\]

∴ \left( \sqrt{10}+\sqrt{8} \right) বড়ো।

(iii) দুটি মিশ্র দ্বিঘাত করণী লিখি যাদের গুণফল একটি মূলদ সংখ্যা/

উত্তর –

মনেকরি, দুটি মিশ্র দ্বিঘাত করণী যথাক্রমে \left( x+\sqrt{y} \right),\,\,\left( x-\sqrt{y} \right)[ যেখানে, x, y মূলদ সংখ্যা]

\left( x+\sqrt{y} \right)\,\left( x-\sqrt{y} \right)={{x}^{2}}-{{\left( \sqrt{y} \right)}^{2}}={{x}^{2}}-y, মূলদ সংখ্যা।

∴নির্নেয় মিশ্র দ্বিঘাত করণী দুটি যথাক্রমে \left( x+\sqrt{y} \right),\,\,\left( x-\sqrt{y} \right)

(iv) \sqrt{72} থেকে কত বিয়োগ করলে \sqrt{32} হবে তা লিখি।

উত্তর –

মনেকরি, x বিয়োগ করতে হবে।

প্রশ্নানুসারে,

\[\sqrt{72}-x=\sqrt{32}\]

\[\Rightarrow 6\sqrt{2}-x=4\sqrt{2}\]

\[\Rightarrow -x=4\sqrt{2}-6\sqrt{2}=-2\sqrt{2}\]

\[\therefore \,\,x=2\sqrt{2}\]

∴\sqrt{72} থেকে 2\sqrt{2}বিয়োগ করলে \sqrt{32} হবে।

(v) \left( \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right) -এর সরলতম মান লিখি।

উত্তর –

\[\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\]

\[=\frac{\left( \sqrt{2}-1 \right)}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}+\frac{\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}+\frac{\left( \sqrt{4}-\sqrt{3} \right)}{\left( \sqrt{4}+\sqrt{3} \right)\left( \sqrt{4}-\sqrt{3} \right)}\]

\[=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\]

\[=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\]

\[=\sqrt{4}-1=2-1=1\]

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