মাধ্যমিক গণিত – দ্বিঘাত করণী – কষে দেখি – 9.3

এই আর্টিকেলে মাধ্যমিক (দশম শ্রেণী) গণিতের নবম অধ্যায়, ‘দ্বিঘাত করণী’ -এর ‘কষে দেখি – 9.3’ বিভাগের সমস্ত সমস্যার সমাধান করে দেওয়া হয়েছে। এই আর্টিকেলটি তোমাদের মাধ্যমিক পরীক্ষার প্রস্তুতিতে বিশেষভাবে সাহায্য করবে।

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

সমাধান –

(i) \(m^{2}+\frac{1}{m^{2}}\)

\(m^{2}+\frac{1}{m^{2}}\)

= \(\left(m+\frac{1}{m}\right)^{2}-2\times m\times\frac{1}{m}\)

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

= \(3-2\)

= \(1\)

(ii) \(m^{3}+\frac{1}{m^{3}}\)

\(m^{3}+\frac{1}{m^{3}}\)

= \(\left(m+\frac{1}{m}\right)^{3}-3\times m\times\frac{1}{m}\left(m+\frac{1}{m}\right)\)

= \(\left(\sqrt{3}\right)^{3}-3\times\left(\sqrt{3}\right)\) [যেহেতু, \(\left(m+\frac{1}{m}\right)=\sqrt{3}\)]

= \(3\sqrt{3}-3\sqrt{3}\)

= \(0\)

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

সমাধান –

L.H.S = \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

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

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

= \(\frac{4\sqrt{15}}{5-3}\)

= \(\frac{4\sqrt{15}}{2}\)

= \(2\sqrt{15}\) = R.H.S

∴ L.H.S = R.H.S [প্রমাণিত]

2. সরল করি

সমাধান –

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

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

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

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

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

= \(\frac{\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}}{\sqrt{3}\times 2}\)

= \(\frac{2\sqrt{2}}{2\sqrt{3}}\)

= \(\frac{\sqrt{2}}{\sqrt{3}}\)

= \(\frac{\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\) [হরের করণী নিরসন করে পাই]

= \(\frac{\sqrt{6}}{3}\)

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

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

হরের করণী নিরসন করে পাই

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

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

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

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

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

= \(0\)

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

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

= \(\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{3\times 3\times 2}}-\frac{\sqrt{3\times 3\times 2}}{3-\sqrt{2\times 2\times 3}}\)

= \(\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-3\sqrt{2}}-\frac{3\sqrt{2}}{3-2\sqrt{3}}\)

হরের করণী নিরসন করে পাই

= \(\frac{4\sqrt{3}}{2-\sqrt{2}}\times\frac{2+\sqrt{2}}{2+\sqrt{2}}-\frac{30}{4\sqrt{3}-3\sqrt{2}}\times\frac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}-\frac{3\sqrt{2}}{3-2\sqrt{3}}\times\frac{3+2\sqrt{3}}{3+2\sqrt{3}}\)

= \(\frac{4\left(2\sqrt{3}+\sqrt{6}\right)}{\left\{\left(2\right)^{2}-\left(\sqrt{2}\right)^{2}\right\}}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{\left\{\left(4\sqrt{3}\right)^{2}-\left(3\sqrt{2}\right)^{2}\right\}}-\frac{3\left(3\sqrt{2}+2\sqrt{6}\right)}{\left\{\left(3\right)^{2}-\left(2\sqrt{3}\right)^{2}\right\}}\) [যেহেতু, \(a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\)]

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

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

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

= \(4\sqrt{6}\)

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

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

হরের করণী নিরসন করে পাই

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

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

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

= \(\frac{3\left(\sqrt{6}-\sqrt{12}\right)}{-3}-\frac{4\left(\sqrt{18}-\sqrt{6}\right)}{4}+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{-1}\)

= \(-\left(\sqrt{6}-\sqrt{12}\right)-\left(\sqrt{18}-\sqrt{6}\right)-\left(\sqrt{12}-\sqrt{18}\right)\)

= \(0\)

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।

সমাধান –

\(\frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)

x, y এবং z এর মান বসিয়ে পাই,

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

হরের করণী নিরসন করে পাই

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

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

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

= \(\frac{3\left(\sqrt{6}-\sqrt{12}\right)}{-3}-\frac{4\left(\sqrt{18}-\sqrt{6}\right)}{4}+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{-1}\)

= \(-\left(\sqrt{6}-\sqrt{12}\right)-\left(\sqrt{18}-\sqrt{6}\right)-\left(\sqrt{12}-\sqrt{18}\right)\)

= \(0\)

4.[ latex]x=\sqrt{7}+\sqrt{6}[/latex] হলে, (i) \(x-\frac{1}{x}\) (ii) \(x+\frac{1}{x}\) (iii) \(x^{2}+\frac{1}{x^{2}}\) এবং (iv) \(x^{3}+\frac{1}{x^{3}}\) -এর সরলতম মানগুলি নির্ণয় করি।

সমাধান –

\(x=\sqrt{7}+\sqrt{6}\)

∴ \(\frac{1}{x}=\frac{1}{\sqrt{7}+\sqrt{6}}\)

= \(\frac{1}{\sqrt{7}+\sqrt{6}}\times\frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}-\sqrt{6}}\) [হরের করণী নিরসন করে পাই]

= \(\frac{\sqrt{7}-\sqrt{6}}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{6}\right)^{2}}\) [যেহেতু, \(a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\)]

= \(\frac{\sqrt{7}-\sqrt{6}}{7-6}\)

= \(\frac{\sqrt{7}-\sqrt{6}}{1}\)

= \(\sqrt{7}-\sqrt{6}\)

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

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

= \(2\sqrt{6}\)

(ii) \(x+\frac{1}{x}\)

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

= \(2\sqrt{7}\)

(iii) \(x^{2}+\frac{1}{x^{2}}\)

= \(\left(x-\frac{1}{x}\right)^{2}+2\times x\times\frac{1}{x}\)

= \(\left(2\sqrt{6}\right)^{2}+2\times 1\)

= \(24+2\)

= \(26\)

(iv) \(x^{3}+\frac{1}{x^{3}}\)

= \(\left(x+\frac{1}{x}\right)^{3}-3\times x\times\frac{1}{x}\left(x+\frac{1}{x}\right)\)

= \(\left(2\sqrt{7}\right)^{3}-3\times 1\times\left(2\sqrt{7}\right)\)

= \(56\sqrt{7}-6\sqrt{7}\)

= \(50\sqrt{7}\)

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।

সমাধান –

\(\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}\)

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

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

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

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

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

= \(4x^{2}-2\)

যেহেতু প্রদত্ত রাশির মান 14

∴ \(4x^{2}-2=14\)

বা, \(4x^{2}=14+2\)

বা, \(4x^{2}=16\)

বা, \(x^{2}=\frac{16}{4}\)

বা, \(x^{2}=4\)

বা, \(x=\pm\sqrt{4}\)

বা, \(x=\pm 2\)

∴ x -এর মান ±2

6. যদি \(a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\) ও \(b=\frac{\sqrt{5}-1}{\sqrt{5}+1}\) হয়, তবে নীচের মানগুলি নির্ণয় করি।

সমাধান –

\(a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\) ও \(b=\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

∴ \(a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

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

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

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

= \(\frac{2\left(5+1\right)}{5-1}\)

= \(\frac{2\times 6}{4}\)

= \(\frac{12}{4}\)

= \(4\)

\(a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\) ও \(b=\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

∴ \(a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

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

= \(\frac{4\times\sqrt{5}\times 1}{\left(\sqrt{5}\right)^{2}-\left(1\right)^{2}}\) [∵ \(\left(a+b\right)^{2}-\left(a-b\right)^{2}=4ab\)]

= \(\frac{4\sqrt{5}}{5-1}\)

= \(\frac{4\sqrt{5}}{4}\)

= \(\sqrt{5}\)

\(a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\) ও \(b=\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

∴ \(a\times b=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times\frac{\sqrt{5}-1}{\sqrt{5}+1}\)

= \(1\)

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

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

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

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

= \(\frac{\left(4\right)^{2}-1}{\left(\sqrt{5}\right)^{2}+1}\) [∵ a + b = 4, a – b = \(\sqrt{5}\) এবং ab = 1]

= \(\frac{16-1}{5+1}\)

= \(\frac{15}{6}\)

= \(\frac{5}{2}\)

= \(2.5\)

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

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

= \(\frac{\left(\sqrt{5}\right)^{3}}{\left(4\right)^{3}}\) [∵ a + b = 4, a – b = \(\sqrt{5}\)]

= \(\frac{5\sqrt{5}}{64}\)

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

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

= \(\frac{3a^{2}+6ab+3b^{2}-ab}{3a^{2}-6ab+3b^{2}+ab}\)

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

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

= \(\frac{3\left(4\right)^{2}-1}{3\left(\sqrt{5}\right)^{2}+1}\)

[∵ a + b = 4, a – b = \(\sqrt{5}\) এবং ab = 1]

= \(\frac{48-1}{15+1}\)

= \(\frac{47}{16}\)

= \(2.9375\)

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

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

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

= \(\frac{\left(4\right)^{3}-3\times 1\times\left(4\right)}{\left(\sqrt{5}\right)^{3}+3\left(1\right)\left(\sqrt{5}\right)}\)

= \(\frac{64-12}{5\sqrt{5}+3\sqrt{5}}\)

= \(\frac{52}{8\sqrt{5}}\)

= \(\frac{13}{2\sqrt{5}}\)

= \(\frac{13\times\sqrt{5}}{2\sqrt{5}\times\sqrt{5}}\) [হরের করণী নিরসন করে পাই]

= \(\frac{13\sqrt{5}}{10}\)

7.যদি \(x=2+\sqrt{3}\), \(y=2-\sqrt{3}\) হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্ণয় করি।

সমাধান –

\(x=2+\sqrt{3}\)

∴ \(\frac{1}{x}=\frac{1}{2+\sqrt{3}}\)

বা, \(\frac{1}{x}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}\) [হরের করণী নিরসন করে পাই]

বা, \(\frac{1}{x}=\frac{2-\sqrt{3}}{\left(2\right)^{2}-\left(\sqrt{3}\right)^{2}}\)

বা, \(\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}\)

বা, \(\frac{1}{x}=2-\sqrt{3}\)

আবার, \(y=2-\sqrt{3}\)

বা, \(\frac{1}{y}=\frac{1}{2-\sqrt{3}}\)

বা, \(\frac{1}{y}=\frac{1}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}\) [হরের করণী নিরসন করে পাই]

বা, \(\frac{1}{y}=\frac{2+\sqrt{3}}{\left(2\right)^{2}-\left(\sqrt{3}\right)^{2}}\)

বা, \(\frac{1}{y}=\frac{2+\sqrt{3}}{4-3}\)

বা, \(\frac{1}{y}=2+\sqrt{3}\)

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

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

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

= \(2\sqrt{3}\)

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

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

= \(\left(y+\frac{1}{y}\right)^{2}-2\times y\times\frac{1}{y}\)

= \(\left(2-\sqrt{3}+2+\sqrt{3}\right)^{2}-2\) [∴ y এবং \(\frac{1}{y}\) এর মান বসিয়ে পাই]

= \(\left(4\right)^{2}-2\)

= \(16-2\)

= \(14\)

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

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

\(\left(x – \frac{1}{x}\right)^{3}
+ 3 \cdot \left(x \cdot \frac{1}{x}\right)\left(x – \frac{1}{x}\right)\)

= \(\left(2\sqrt{3}\right)^{3}+3\left(2\sqrt{3}\right)\) [\(\left(x-\frac{1}{x}\right)\) এর মান বসিয়ে পাই]

= \(24\sqrt{3}+6\sqrt{3}\)

= \(30\sqrt{3}\)

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

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

\(x=2+\sqrt{3}\) এবং \(y=2-\sqrt{3}\)

∴ x × y

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

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

= \(4-3\)

= \(1\)

∴ \(xy+\frac{1}{xy}\)

= \(1+\frac{1}{1}\)

= \(2\)

(b) \(3x^{2}-5xy+3y^{2}\)

\(x=2+\sqrt{3}\) এবং \(y=2-\sqrt{3}\)

∴ x × y

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

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

= \(4-3\)

= \(1\)

এবং (x – y)

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

= \(2\sqrt{3}\)

∴ \(3x^{2}-5xy+3y^{2}\)

= \(3x^{2}-6xy+3y^{2}+xy\)

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

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

= \(3\left(2\sqrt{3}\right)^{2}+1\) [(x – y) এবং xy এর মান বসিয়ে পাই]

= \(3\times 12+1\)

= \(36+1\)

= \(37\)

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

সমাধান –

\(x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}\)

এবং xy = 1

∴ \(y=\frac{1}{x}\)

বা, \(y=\frac{1}{\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}}\)

বা, \(y=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)

∴ x + y

= \(\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)

= \(\frac{\left(\sqrt{7}+\sqrt{3}\right)^{2}+\left(\sqrt{7}-\sqrt{3}\right)^{2}}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}\)

= \(\frac{2\left\{\left(\sqrt{7}\right)^{2}+\left(\sqrt{3}\right)^{2}\right\}}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}\) [∵ \(\left(a+b\right)^{2}+\left(a-b\right)^{2}=2\left(a^{2}+b^{2}\right)\)]

= \(\frac{2\left(7+3\right)}{7-3}\)

= \(\frac{2\times 10}{4}\)

= \(\frac{20}{4}\)

= \(5\)

∴ \(\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}\)

= \(\frac{x^{2}+2xy+y^{2}-xy}{x^{2}+2xy+y^{2}-3xy}\)

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

= \(\frac{\left(5\right)^{2}-1}{\left(5\right)^{2}-3\times 1}\)

= \(\frac{25-1}{25-3}\)

= \(\frac{24}{22}\)

= \(\frac{12}{11}\) [প্রমাণিত]

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

সমাধান –

\(\left(\sqrt{7}+1\right)^{2}\)

= \(\left(\sqrt{7}\right)^{2}+2\times\sqrt{7}\times 1+\left(1\right)^{2}\)

= \(7+2\sqrt{7}+1\)

= \(8+2\sqrt{7}\)

আবার

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

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

= \(5+2\sqrt{15}+3\)

= \(8+2\sqrt{15}\)

এখন, \(8+2\sqrt{15}>8+2\sqrt{7}\) [∵ \(\sqrt{15}>\sqrt{7}\Rightarrow 2\sqrt{15}>2\sqrt{7}\)]

∴ \(\left(\sqrt{5}+\sqrt{3}\right)^{2}>\left(\sqrt{7}+1\right)^{2}\)

∴ \(\left(\sqrt{5}+\sqrt{3}\right)>\left(\sqrt{7}+1\right)\)

10. অতিসংক্ষিপ্ত 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) \(2\sqrt{3}\)
(c) 4
(d) \(2-\sqrt{3}\)

উত্তর – (c) 4

সমাধান,

\(x=2+\sqrt{3}\)

∴ \(\frac{1}{x}=\frac{1}{2+\sqrt{3}}\)

বা, \(\frac{1}{x}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}\)

বা, \(\frac{1}{x}=\frac{2-\sqrt{3}}{\left(2\right)^{2}-\left(\sqrt{3}\right)^{2}}\)

বা, \(\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}\)

বা, \(\frac{1}{x}=2-\sqrt{3}\)

∴ \(x+\frac{1}{x}\)

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

= \(4\)

∴ \(x+\frac{1}{x}=4\)

(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 – (a) 2

সমাধান,

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

= \(\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}\) [∵ \(p+q=\sqrt{13}\) এবং \(p-q=\sqrt{5}\)]

= \(\frac{13}{4}-\frac{5}{4}\)

= \(\frac{13-5}{4}\)

= \(\frac{8}{4}\)

= \(2\)

(iii) যদি \(a+b=\sqrt{5}\) এবং \(a-b=\sqrt{3}\) হয়, তাহলে \(\left(a^{2}+b^{2}\right)\) -এর মান

(a) 8
(b) 4
(c) 2
(d) 1

উত্তর – (b) 4

সমাধান,

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

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

= \(\frac{5}{4}-\frac{3}{4}\)

= \(\frac{5-3}{4}\)

= \(\frac{2}{4}\)

= \(\frac{1}{2}\)

∴ (a2 + b2)

= (a + b)2 – 2ab

= \(\left(\sqrt{5}\right)^{2}-2\times\frac{1}{2}\)

= \(5-1\)

= \(4\)

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

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

উত্তর – (a) \(\sqrt{80}\)

সমাধান,

\(\sqrt{125}-\sqrt{5}\)

= \(\sqrt{5\times 5\times 5}-\sqrt{5}\)

= \(5\sqrt{5}-\sqrt{5}\)

= \(4\sqrt{5}\)

= \(\sqrt{5\times 4\times 4}\)

= \(\sqrt{80}\)

(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 – (b) 44

সমাধান,

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

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

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

= \(\left(25-3\right)\left(3-1\right)\)

= \(22\times 2\)

= \(44\)

(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি –

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

উত্তর – সত্য।

সমাধান,

\(\sqrt{75}=\sqrt{5\times 5\times 3}=5\sqrt{3}\)

\(\sqrt{147}=\sqrt{7\times 7\times 3}=7\sqrt{3}\)

অর্থাৎ প্রদত্ত করণীদুটি সদৃশ।

(ii) \(\sqrt{\pi}\) একটি দ্বিঘাত করণী।

উত্তর – মিথ্যা।

সমাধান,

\(\sqrt{\pi}\) একটি দ্বিঘাত করণী নয়

অর্থাৎ বিবৃতি 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।

11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.)

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

সমাধান –

\(x=3+2\sqrt{2}\)

বা, \(\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\)

বা, \(\frac{1}{x}=\frac{3-2\sqrt{2}}{\left(3\right)^{2}-\left(2\sqrt{2}\right)^{2}}\)

বা, \(\frac{1}{x}=\frac{3-2\sqrt{2}}{9-8}\)

বা, \(\frac{1}{x}=3-2\sqrt{2}\)

∴ \(x+\frac{1}{x}\)

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

= \(6\)

∴ \(x+\frac{1}{x}=6\)

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

সমাধান –

\(\left(\sqrt{15}+\sqrt{3}\right)^{2}\)

= \(\left(\sqrt{15}\right)^{2}+2\times\sqrt{15}\times\sqrt{3}+\left(\sqrt{3}\right)^{2}\)

= \(15+2\sqrt{45}+3\)

= \(18+2\sqrt{45}\)

এবং \(\left(\sqrt{10}+\sqrt{8}\right)^{2}\)

= \(\left(\sqrt{10}\right)^{2}+2\times\sqrt{10}\times\sqrt{8}+\left(\sqrt{8}\right)^{2}\)

= \(10+2\sqrt{80}+8\)

= \(18+2\sqrt{80}\)

এখন, \(18+2\sqrt{80}>18+2\sqrt{45}\) [যেহেতু, \(\sqrt{80}>\sqrt{45}\)]

∴ \(\left(\sqrt{10}+\sqrt{8}\right)^{2}>\left(\sqrt{15}+\sqrt{3}\right)^{2}\)

∴ \(\left(\sqrt{10}+\sqrt{8}\right)>\left(\sqrt{15}+\sqrt{3}\right)\)

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

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

\(\left(\sqrt{3}+2\right)\) এবং \(\left(\sqrt{3}-2\right)\) হল দুটি মিশ্র দ্বিঘাত করণী যাদের গুনফল একটি মূলদ সংখ্যা।

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

সমাধান –

ধরি \(\sqrt{72}\) থেকে x বিয়োগ করলে \(\sqrt{32}\) হবে।

∴ \(\sqrt{72}-x=\sqrt{32}\)

বা, \(x=\sqrt{72}-\sqrt{32}\)

বা, \(x=\sqrt{2\times 2\times 2\times 3\times 3}-\sqrt{2\times 2\times 2\times 2\times 2}\)

বা, \(x=2\times 3\sqrt{2}-2\times 2\sqrt{2}\)

বা, \(x=6\sqrt{2}-4\sqrt{2}\)

বা, \(x=2\sqrt{2}\)

∴ \(\sqrt{72}\) 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।

সমাধান –

\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)

হরের করণী নিরসন করে পাই

= \(\frac{1}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\times\frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}\)

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

= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\)

= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)

= \(-1+2\)

= \(1\)

---

এই আর্টিকেলে মাধ্যমিক (দশম শ্রেণী) গণিতের নবম অধ্যায়, ‘দ্বিঘাত করণী’ -এর ‘কষে দেখি – 9.3’ বিভাগের সমস্ত সমস্যার সমাধান করা হয়েছে।

আশা করি, এই আর্টিকেলটি আপনাদের পরীক্ষার প্রস্তুতিতে কিছুটা হলেও সহায়ক হয়েছে। যদি কোনো প্রশ্ন, মতামত বা সাহায্যের প্রয়োজন হয়, নিচে কমেন্ট করে জানাতে পারেন অথবা টেলিগ্রামের মাধ্যমে যোগাযোগ করতে পারেন—আমরা আপনাদের সকল প্রশ্নের উত্তর দেওয়ার জন্য সর্বদা প্রস্তুত।