Rút gọn các biểu thức sau: a) A = [1/a^2 + 1/b^2 + 2/(a+b).(1/a + 1/b)].ab/(a + b)^2

a) $A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\cdot \left(\frac{1}{a}+\frac{1}{b}\right)\right]\cdot \frac{ab}{{\left(a+b\right)}^2}$

$=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right]\cdot \frac{ab}{{\left(a+b\right)}^2}$$=\frac{a^2+b^2+2ab}{a^2{b}^2}\cdot \frac{ab}{{\left(a+b\right)}^2}$

$=\frac{{\left(a+b\right)}^2}{a^2{b}^2}\cdot \frac{ab}{{\left(a+b\right)}^2}=\frac{1}{ab}.$

b) $B=\left[\frac{1}{{\left(2x-y\right)}^2}+\frac{2}{4x^2-y^2}+\frac{1}{{\left(2x+y\right)}^2}\right]\cdot \frac{4x^2+4xy+y^2}{16x}$

$=\left[\frac{1}{{\left(2x-y\right)}^2}+\frac{2}{\left(2x+y\right)\left(2x-y\right)}+\frac{1}{{\left(2x+y\right)}^2}\right]\cdot \frac{{\left(2x+y\right)}^2}{16x}$

$=\frac{{\left(2x-y\right)}^2+2\left(2x+y\right)\left(2x-y\right)+{\left(2x+y\right)}^2}{{\left(2x+y\right)}^2{\left(2x-y\right)}^2}\cdot \frac{{\left(2x+y\right)}^2}{16x}$

$=\frac{{\left[\left(2x-y\right)+\left(2x+y\right)\right]}^2}{{\left(2x-y\right)}^2}\cdot \frac{1}{16x}$$=\frac{16x^2}{16x\cdot {\left(2x-y\right)}^2}=\frac{x}{{\left(2x-y\right)}^2}$