If $x + \frac{1}{x} = 5, x ≠ 0 $ then the value of $\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$ is equal to : |
55 60 65 50 |
55 |
$x + \frac{1}{x} = 5, x ≠ 0 $ then the value of $\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$ We know that, If x + \(\frac{1}{x}\) = n then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n Divide numerator and denominator by x we get, We can write $\frac{x^3+\frac{1}{x^3}}{x-3+\frac{1}{x}}$ = $x^3 +\frac{1}{x^3}$ = 53 - 3 × 5 = 110 Put this value in $\frac{x^3+\frac{1}{x^3}}{x-3+\frac{1}{x}}$ = $\frac{110}{5-3}$ = 55 |