Question

If $x-\frac{1}{x}=3$; then the value of $\frac{3x^2-3}{x^2+2x-1}$ is

• A. 9/5
• B. 8/5
• C. 7/5
• D. 6/5

Answer 1

The correct option is A 9/5

Given $x+\frac{1}{x}=3$ multiply by x both sides

Then $x^2-1=3x$

$\Rightarrow x^2-3x-1=0$

So $\frac{3x^2-3}{x^2+2x-1}=\frac{3(x^2-1)}{x^2-3x-1+5x}=\frac{3\times 3x}{0+5x}=\frac{9x}{5x}=\frac{9}{5}$