x² = 5 + 2√6 হলে, $\frac{x^{6} + 1}{x^{3}} - \sqrt{2} (\frac{x^{4} + 1}{x^{2}})$ এর মান কত?

dsuc.created: 2 months ago | dsuc.updated: 2 months ago

dsuc.updated: 2 months ago

দেওয়া আছে, $x^{2} = 5 + 2 \sqrt{6}$

∴ $\frac{1}{x^{2}} = \frac{1}{5 + 2 \sqrt{6}}$

$= \frac{5 - 2 \sqrt{6}}{(5 + 2 \sqrt{6}) (5 - 2 \sqrt{6})}$

$= \frac{5 - 2 \sqrt{6}}{5^{2} - {(2 \sqrt{6})}^{2}}$

$= \frac{5 - 2 \sqrt{6}}{25 - 24}$

$= 5 - 2 \sqrt{6}$

∴ $x^{2} + \frac{1}{x^{2}} = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} = 10$

প্রদত্ত রাশি $= \frac{x^{6} + 1}{x^{3}} - \sqrt{2} (\frac{x^{4} + 1}{x^{2}})$

$= x^{3} + \frac{1}{x^{3}} - \sqrt{2} (x^{2} + \frac{1}{x^{2}})$

$= {(x + \frac{1}{x})}^{3} - 3 . x . \frac{1}{x} (x + \frac{1}{x}) - \sqrt{2} \times 10$

$= {(2 \sqrt{3})}^{3} - 3 \times 2 \sqrt{3} - 10 \sqrt{2}$

$= 18 \sqrt{3} - 10 \sqrt{2}$ (Answer)

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