<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] $${ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( x-1 \right)  }^{ 2 } } =2010\\ { \left( x+\frac { x }{ x-1 }  \right)  }^{ 2 }-2\frac { { x }^{ 2 } }{ x-1 } =2010\\ { \left( \frac { { x }^{ 2 } }{ x-1 }  \right)  }^{ 2 }-2\left( \frac { { x }^{ 2 } }{ x-1 }  \right) -2010=0\\ \left( \frac { { x }^{ 2 } }{ x-1 }  \right) =t\\ { t }^{ 2 }-2t-2010=0\\ t=1\pm \sqrt { 2011 } \\ \\ $$
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