How do you solve x^2/(x^2-4) = x/(x+2)-2/(2-x)x2x2−4=xx+2−22−x?

1 Answer
Anish M.
Mar 24, 2018

There is no solution

Explanation:

x^2/(x^2-4)=x/(x+2)-2/(2-x)x2x2−4=xx+2−22−x
Becomes x^2/(x^2-4)=x/(x+2)+2/(x-2)x2x2−4=xx+2+2x−2

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

Becomes x^2/(x^2-4)=(x(x-2))/((x+2)(x-2))+(2(x+2))/((x-2)(x+2))x2x2−4=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)x2x2−4=x2−2x+2x+4x2−4

Becomes x^2/(x^2-4)=(x^2 + 4)/(x^2-4)x2x2−4=x2+4x2−4

Becomes x^2=(x^2 + 4)x2=(x2+4)

There is no solution

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