1) যদি \displaystyle x=3+2\sqrt{2} হলে,

ক) \displaystyle \frac{1}{x} এর মান নির্নয় কর ।

খ) \displaystyle x^{6} +\frac{1}{x^{6}} এর মান নির্নয় কর ।

গ) প্রমান কর যে, \displaystyle \left(\sqrt{x}\right)^{3} -\left(\frac{1}{\sqrt{x}}\right)^{3} =14

ক)

দেওয়া আছে,

\displaystyle x=3+2\sqrt{2}

\displaystyle \frac{1}{x} =\frac{1}{3+2\sqrt{2}}

\displaystyle =\frac{1.\left( 3-2\sqrt{2}\right)}{\left( 3+2\sqrt{2}\right) .\left( 3-2\sqrt{2}\right)} [উভয় পক্ষ \displaystyle 3-2\sqrt{2} দ্বারা ভাগ করে পাই ]

\displaystyle =\frac{\left( 3-2\sqrt{2}\right)}{( 3)^{2} -\left( 2\sqrt{2}\right)^{2}} [ \displaystyle a^{2} -b^{2} =( a+b)( a-b) সূত্র প্রয়োগ করে পাই ]

\displaystyle =\frac{\left( 3-2\sqrt{2}\right)}{( 9)^{2} -( 2)^{2} .\left(\sqrt{2}\right)^{2}}

\displaystyle =\frac{\left( 3-2\sqrt{2}\right)}{( 9-4.2)}

\displaystyle =\frac{\left( 3-2\sqrt{2}\right)}{9-8}

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

\displaystyle =3-2\sqrt{2}

নির্নয় মান: \displaystyle 3-2\sqrt{2}

খ)

দেওয়া আছে,

\displaystyle x=3+2\sqrt{2}

(ক) হতে পাই,

\displaystyle \frac{1}{x} =3-2\sqrt{2} \

\displaystyle x+\frac{1}{x} =3+2\sqrt{2} +3-2\sqrt{2}

\displaystyle =6
এখন,
\displaystyle x^{2} +\frac{1}{x^{2}} =\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x}

\displaystyle =( 6)^{2} -2

\displaystyle =36-2

\displaystyle =34

সূতরাং, \displaystyle a^{6} +\frac{1}{a^{6}} =\left( x^{2}\right)^{3} +\left(\frac{1}{x^{2}}\right)^{3}

\displaystyle =\left( x^{2} +\frac{1}{x^{2}}\right)^{3} -3.x^{2} .\frac{1}{x^{2}}\left( x^{2} +\frac{1}{x^{2}}\right)

\displaystyle =( 34)^{3} -3.34

\displaystyle =34.34.34-102

\displaystyle =39304-102

\displaystyle =39202

গ)


দেওয়া আছে,
\displaystyle x=3+2\sqrt{2}

\displaystyle \Longrightarrow x=2+2\sqrt{2} +1

\displaystyle \Longrightarrow x=\left(\sqrt{2}\right)^{2} +2.1.\sqrt{2} +1

\displaystyle \Longrightarrow x=\left(\sqrt{2} +1\right)^{2}

\displaystyle \Longrightarrow \sqrt{x} =\sqrt{\left(\sqrt{2} +1\right)^{2}}

\displaystyle \Longrightarrow \sqrt{x} =\left(\sqrt{2} +1\right)

\displaystyle \Longrightarrow \frac{1}{\sqrt{x}} =\frac{1}{\left(\sqrt{2} +1\right)}

\displaystyle \Longrightarrow \frac{1}{\sqrt{x}} =\frac{\ \left(\sqrt{2} -1\right)}{\left(\sqrt{2} +1\right) .\left(\sqrt{2} -1\right)}

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

\displaystyle \Longrightarrow \frac{1}{\sqrt{x}} =\frac{\ \left(\sqrt{2} -1\right)}{2-1}

\displaystyle \Longrightarrow \frac{1}{\sqrt{x}} =\frac{\ \left(\sqrt{2} -1\right)}{1}

\displaystyle \Longrightarrow \frac{1}{\sqrt{x}}=\displaystyle \left(\sqrt{2} -1\right)

সুতরাং,

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

\displaystyle =2

এখন, \displaystyle \left(\sqrt{x}\right)^{3} -\left(\frac{1}{\sqrt{x}}\right)^{3}

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

\displaystyle =( 2)^{2} +3.2

\displaystyle =2.2.2+6

\displaystyle =8+6

\displaystyle =14

সুতরাং, \displaystyle \left(\sqrt{x}\right)^{3} -\left(\frac{1}{\sqrt{x}}\right)^{3} =14 ( প্রমানিত )

2) যদি, \displaystyle x=\sqrt{13+2\sqrt{42}} হয়। তবে,

ক) \displaystyle x+\frac{1}{x} এর মান নির্নয় কর ।

খ) প্রমান কর যে, \displaystyle x^{3} -\frac{1}{x^{3}} =54\sqrt{6}

গ) \displaystyle x^{5} +\frac{1}{x^{5}} এর মান নির্নয় কর ।

ক)

দেওয়া আছে,

\displaystyle x=\sqrt{13+2\sqrt{42}}

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{13+2\sqrt{42}}\right)^{2} [ বর্গ করে পাই ]

\displaystyle \Longrightarrow x^{2} =13+2\sqrt{42}

\displaystyle \Longrightarrow x^{2} =7+2\sqrt{42} +6

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{7}\right)^{2} +2.\sqrt{7} .\sqrt{6} +\left(\sqrt{6}\right)^{2}

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{7} +\sqrt{6}\right)^{2}

\displaystyle \Longrightarrow x=\sqrt{7} +\sqrt{6}

আবার, \displaystyle \frac{1}{x} =\frac{1}{\sqrt{7} +\sqrt{6}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1.\left(\sqrt{7} -\sqrt{6}\right)}{\left(\sqrt{7} +\sqrt{6}\right) .\left(\sqrt{7} -\sqrt{6}\right)}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} -\sqrt{6}\right)}{\left(\sqrt{7}\right)^{2} -\left(\sqrt{6}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} -\sqrt{6}\right)}{7-6}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} -\sqrt{6}\right)}{1}

\displaystyle \Longrightarrow \frac{1}{x} =\left(\sqrt{7} -\sqrt{6}\right)

এখন,
\displaystyle x+\frac{1}{x} =\sqrt{7} +\sqrt{6} +\sqrt{7} -\sqrt{6}

\displaystyle =2\sqrt{7}

খ)

(ক) হতে পাই,

\displaystyle x=\sqrt{7} +\sqrt{6}

এবং, \displaystyle \frac{1}{x} =\sqrt{7} -\sqrt{6}

এখানে, \displaystyle x-\frac{1}{x} =\sqrt{7} +\sqrt{6} -\sqrt{7} +\sqrt{6}

\displaystyle =2\sqrt{6}

বামপক্ষ\displaystyle =x^{3} -\frac{1}{x^{3}}

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

\displaystyle =\left( 2\sqrt{6}\right)^{3} +3.2\sqrt{6}

\displaystyle =( 2)^{3} .\left(\sqrt{6}\right)^{3} +6\sqrt{6}

\displaystyle =2.2.2.\sqrt{6.}\sqrt{6} .\sqrt{6} +6\sqrt{6}

\displaystyle =8.\left(\sqrt{6}\right)^{2} .\sqrt{6} +6\sqrt{6}

\displaystyle =8.6\sqrt{6} +6\sqrt{6}

\displaystyle =48\sqrt{6} +6\sqrt{6}

\displaystyle =54\sqrt{6}

\displaystyle =ডানপক্ষ

সুতরাং, \displaystyle x^{3} -\frac{1}{x^{3}} =54\sqrt{6} ( প্রমানিত )

গ)

(ক) হতে পাই,

\displaystyle x+\frac{1}{x} =2\sqrt{7}

এখানে,
\displaystyle x^{2} +\frac{1}{x^{2}} =\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x}

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

\displaystyle =( 2)^{2} .\left(\sqrt{7}\right)^{2} -2

\displaystyle =\ 4.7-2

\displaystyle =28-2

\displaystyle =26

আবার,
\displaystyle x^{3} +\frac{1}{x^{3}} =\left( x+\frac{1}{x}\right)^{3} -3.x.\frac{1}{x}\left( x+\frac{1}{x}\right)

\displaystyle =\left( 2\sqrt{7}\right)^{3} -3.2\sqrt{7}

\displaystyle =( 2)^{3} .\left(\sqrt{7} .\sqrt{7} .\sqrt{7}\right) -6\sqrt{7}

\displaystyle =2.2.2.\left(\sqrt{7}\right)^{2} .\sqrt{7} -6\sqrt{7}

\displaystyle =8.7\sqrt{7} -6\sqrt{7}

\displaystyle =56\sqrt{7} -6\sqrt{7}

\displaystyle =50\sqrt{7}

\displaystyle \Longrightarrow 1300\sqrt{7} =x^{5} +\frac{1}{x^{5}} +2\sqrt{7}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =1300\sqrt{7} -2\sqrt{7}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =1298\sqrt{7}

3) \displaystyle x=1+\sqrt{3} এবং \displaystyle y=\sqrt{3} -1

ক) \displaystyle ( x-1) এবং \displaystyle x এর মাঝে একটি মূলদ সংখ্যা নির্নয় কর ।

খ) \displaystyle \frac{x^{4}}{y^{4}} -\frac{y^{4}}{x^{4}} এর মান নির্নয় কর ।

গ) প্রমান কর যে, \displaystyle \frac{\left( x^{4} +4\right)\left( x^{6} -8\right)}{x^{3}} =160

Post Author: showrob

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