How do you solve $x^2/(x^2-4) = x/(x+2)-2/(2-x)$ ?

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Answer 1 · 2018-03-24T19:09:28

There is no solution

$x^2/(x^2-4)=x/(x+2)-2/(2-x)$
Becomes $x^2/(x^2-4)=x/(x+2)+2/(x-2)$

On the right side, multiply and divide first fraction with $x-2$
On the right side, multiply and divide second fraction with $x+2$
We get,

Becomes $x^2/(x^2-4)=(x(x-2))/((x+2)(x-2))+(2(x+2))/((x-2)(x+2))$

Becomes $x^2/(x^2-4)=(x^2-2x + 2x + 4)/(x^2-4)$

Becomes $x^2/(x^2-4)=(x^2 + 4)/(x^2-4)$

Becomes $x^2=(x^2 + 4)$

There is no solution

How do you solve x^2/(x^2-4) = x/(x+2)-2/(2-x) x^2/(x^2-4) = x/(x+2)-2/(2-x)?