Question

please answer it fastly​

Answer 1

Answer:

Given

x+$\frac{1}{4}$= 3

Squaring on both sides

x+$\frac{1}{x}$^2= $3^2$

$\implies$$x^2$ + 2 × x × $\frac{1}{x}$ +($\frac{1}{x}$)^2 = 9

$\implies$$x^2$ + 2 +$\frac{1}{x^2}$ = 9

$x^2$ + $\frac{1}{x^2}$= 7

Again squaring both sides

( $x^2$ + $\frac{1}{x^2}$ )^2 = $7^2$

$\implies$ $(x^2)^2$ + 2 × $x^2$ × $\frac{1}{x^2}$ + ($\frac{1}{x^2}$)^2 = 49

$\implies$$x^4$ + 2 + $\frac{1}{x^4}$ = 49

$x^4$+$\frac{1}{x^4}$ = 47 Ans.

Answer 2

Answer:

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