`a)A=[(2x^2+2x)(x-2)^2]/[(x^3-4x)(x+1)]`
`A=[2x(x+1)(x-2)^2]/[x(x-2)(x+2)(x+1)]`
`A=[2x-4]/[x+2]`
Thay `x=1/2` vào `A` có: `A=[2. 1/2-4]/[1/2+2]=-6/5`
`b)B=[x^3-x^2y+xy^2]/[x^3+y^3]`
`B=[x(x^2-xy+y^2)]/[(x+y)(x^2-xy+y^2)]`
`B=x/[x+y]`
Thay `x=-5;y=10` vào `B` có: `B=[-5]/[-5+10]=-1`
a: \(=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}=\dfrac{2\cdot\dfrac{1}{2}-2}{\dfrac{1}{2}+2}=-1:\dfrac{5}{2}=-\dfrac{2}{5}\)
b: \(=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)