<s>[INST] $ 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
 [/INST] A straightforward approach not requiring recognition of an identity runs:
Let $y=\frac {x^2}{x-1}$ so that $x^2=y(x-1)$, which means we can reduce $x^2$ wherever it occurs and see what happens. So $$x^2+\frac {x^2}{(x-1)^2}=y\left(x-1+\frac 1{x-1}\right)=y\cdot\frac {x^2-2x+2}{x-1}=y^2-2y$$And it is easy from there (solve the quadratic)
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