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

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#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