#### Answer

No solution.

#### Work Step by Step

$\displaystyle \frac{1}{x-1}-\frac{2}{x^{2}}=0$
$\displaystyle \frac{1}{x-1}=\frac{2}{x^{2}}$
We cross-multiply:
$x^{2}=2(x-1)$
$x^{2}-2(x-1)=0$
$x^{2}-2x+2=0$
Next we use the quadratic formula:
$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4*1*2}}{2(1)}=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm\sqrt{-4}}{2}$
Since the discriminant is negative, there are no real solutions.