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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

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  • $\begingroup$ Since they don't ask about $x$, it might be easier to find $\frac{x^2}{x-1}$ directly. $\endgroup$ – Arthur Jul 25 '15 at 7:59
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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, $\frac{x^2}{x-1} = 1\pm \sqrt{2011}$

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

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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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