Question

A= \(\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)

Answer 1

\(A=\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)

\(A=\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{x-4}\)

\(A=\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\dfrac{\sqrt{x}-2-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\dfrac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\dfrac{-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{-6}{x-4}\)