Question

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

• A. 5/4
• B. 1/4
• C. 1/3
• D. 3/4

Answer 1

The correct option is B 1/4

Given $x+\frac{1}{x}=1$

Then $\frac{x}{x^2+x+1}$

If it written as

$\frac{x^2+x+1}{x}=x+1+\frac{1}{x}=x+\frac{1}{x}+1=3+1=4$

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