\(x = \sqrt 3 + \sqrt 2 \) হলে, \(x2 + \frac{1}{{x2}}\) -এর মান নির্ণয় করুন
Correct Answer: Option A
দেওয়া আছে , \(x = \sqrt 3 + \sqrt 2 \)
বা, \(\frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }}\)
বা, \(\frac{1}{x} = \frac{{1(\sqrt 3 - \sqrt 2 )}}{{(\sqrt 3 + \sqrt 2 )(\sqrt 3 - \sqrt 2 )}}\)
বা, \(\frac{1}{x} = \frac{{(\sqrt 3 - \sqrt 2 )}}{{(\sqrt {3)} 2 - (\sqrt 2 )2}}\)
\(\frac{1}{x} = (\sqrt 3 - \sqrt 2 )\)
সুতরাং \(x2 + \frac{1}{{x2}} = (x + \frac{1}{x})2 - 2.x.\frac{1}{x}\)
\( = (\sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 )2 - 2 = (2\sqrt 3 )2 - 2 = (2\sqrt 3 )2 - 2 = 4.3 - 2 = 12 - 2 = 10\)
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