Question

Answer 1

\(=\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x+2\right)}\)

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

\(=\dfrac{x^2+4x-x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+1}{x^2-2x}\)

Answer 2

\(\Leftrightarrow\)\(\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{x\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x^2+4x-x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+3x-2}{x\left(x-2\right)\left(x+2\right)}\)