$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$ 
find $\dfrac {x^2} {x-1}.$ 
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $ (D) $\sqrt {2011}$
So I multiply them with $(x-1)$ 
$$x^2(x-1) + \frac {x^2} {x-1} = 2010(x-1)$$
$$\frac {x^2} {x-1} = (x-1)(2010-x^2)$$
and I stuck in here, dont know how to remove $x$ in there

Hint:
$$a^2 + b^2 = (a+b)^2-2ab$$
Going forward,
$$ \left(x + \frac{x}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
$$ \left(\frac{x^2}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
Let $\frac{x^2}{x-1} = n$
$$ n^2 -2n -2010 =0  $$
You should be able to solve this using the quadratic formula or otherwise to get a value for $ \frac{x^2}{x-1}$
Indeed, $\frac{x^2}{x-1} = 1\pm \sqrt{2011}$
