If $ x + \frac{1}{x} = 10, $ then find the value of $\frac{7x}{x^2 + 1- 8x}.$
Answer & explanation
Correct answer: option 1
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