Question

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

b,\(\dfrac{2}{xy}\):(\(\dfrac{1}{x}\)-\(\dfrac{1}{y}\))-\(\dfrac{x^2-y^2}{\left(x-y\right)^2}\)

c,(\(\dfrac{x+y}{2x-2y}\)-\(\dfrac{x-y}{2x+2y}\)-\(\dfrac{2y^2}{y^2-x^2}\)):\(\dfrac{2y}{x-y}\)

Answer 1

a: \(=\dfrac{2x+2-2}{\left(x-2\right)\left(x+1\right)}\cdot\left(1+x\right)\)

=2x/x-2

b: \(=\dfrac{2}{xy}:\dfrac{y-x}{xy}-\dfrac{x+y}{x-y}\)

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

c: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{x-y}{2y}\)

\(=\dfrac{4y^2+4xy}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)