If $ x + \frac{1}{x} = 10, $ then find the value of $\frac{7x}{x^2 + 1- 8x}.$

3.5

If $ x + \frac{1}{x} = 10, $

then find the value of $\frac{7x}{x^2 + 1- 8x}.$

$ x + \frac{1}{x} = 10, $ = \(\frac{x^2 + 1}{x}\) = 10

= x² + 1 = 10x

Now put this value in $\frac{7x}{x^2 + 1- 8x}$ = $\frac{7x}{10x - 8x}$

$\frac{7x}{x^2 + 1- 8x}$ = \(\frac{7}{2}\) = 3.5