Question

a) $\frac{1}{x-y}$ + $\frac{2}{x+y}$ + $\frac{3x}{y^2-x^2}$ b) $\frac{1}{x-2}$ + $\frac{1}{x+2}$ - $\frac{4x-4}{x^2-4}$ c) $\frac{x+1}{x+3}$ - $\frac{x-1}{3-x}$ + $\frac{2x-2x^2}{x^2-9}$ d) $\frac{1}{3x-2}$ - $\frac{1}{3x+2}$ - $\frac{3x-6}{4-9{}^{}}$

Answer 1

a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)

\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)

b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)

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

=-2/x+2

c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)

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

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

\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)