BD Merit - Freelancing, BCS, Bank, Primary, Admission, NTRCA

Discuss Forum

1. $x^{2} - \sqrt{2 x} + 1 = 0$

Answer: Option A

Explanation:

দেওয়া আছে, x^{2} - \sqrt{2} x + 1 = 0 ≫ x^{2} + 1 = \sqrt{2} x ≫ \frac{x^{2}}{x} + \frac{1}{x} = \frac{\sqrt{2} x}{x} ≫ x + \frac{1}{x} = \sqrt{2} প্রদত্ত রাশি, \frac{x^{12} + 1}{x^{6}} = \frac{x^{12}}{x^{6}} + \frac{1}{x^{6}} = x^{6} + \frac{1}{x^{6}} = {(x^{2})}^{3} + {(\frac{1}{x^{2}})}^{3} = {\{{(x + \frac{1}{x})}^{2} - 2 . x . \frac{1}{x}\}}^{3} - 3 \{{(x + \frac{1}{x})}^{2} - 2 . x . \frac{1}{x}\} = {\{{(\sqrt{2})}^{2} - 2\}}^{3} - 3 \{{(\sqrt{2})}^{2} - 2\} = {(0)}^{3} - 3 x 0 = 0

Post your comments here:

Name:

Mobile:

Email:(Optional)

» Your comments will be displayed only after manual approval.