Làm các phép tính sau:

a) $\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right);$            b) $\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)$.

Giải

a) $\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right)=\frac{x^2.x+y.y^2}{x{y}^2}:\frac{x.x-xy+y^2}{x{y}^2}$

$$\begin{gathered} =\frac{x^3+y^3}{x{y}^2}:\frac{x^2-xy+y^2}{x{y}^2}=\frac{x^3+y^3}{x{y}^2}.\frac{x{y}^2}{x^2-xy+y^2} \\ =\frac{\left(x+y\right)\left(x^2-xy+y^2\right)x{y}^2}{x{y}^2\left(x^2-xy+y^2\right)}=x+y \end{gathered}$$

b) $\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)=\left[\frac{1}{{\left(x+2\right)}^2}-\frac{1}{{\left(x+2\right)}^2}\right]:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}$

$$\begin{gathered} =\frac{{\left(x-2\right)}^2-{\left(x+2\right)}^2}{{\left(x+2\right)}^2{\left(x-2\right)}^2}.\frac{\left(x-2\right)\left(x+2\right)}{2x}=\frac{\left(x^2-4x+4-x^2-4x-4\right)\left(x-2\right)\left(x+2\right)}{2x{\left(x+2\right)}^2{\left(x-2\right)}^2} \\ =\frac{-8x}{2x\left(x+2\right)\left(x-2\right)}=-\frac{4}{\left(x+2\right)\left(x-2\right)}=-\frac{4}{x^2-4}. \end{gathered}$$