a: \(=\dfrac{2\left(x+1\right)\cdot\left(x+1\right)}{2\left(x+1\right)\cdot2x}=\dfrac{x+1}{2x}\)
b: \(=\dfrac{\left(x-8\right)\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x-8}{x+2}\)
\(a,đk:x\ne0;x\ne-1\)
\(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}=\dfrac{2\left(x+1\right)\left(x+1\right)}{4x\left(x+1\right)}=\dfrac{x+1}{2x}\)
\(c,ĐKx\ne-2\\\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x-8}{x+2}\)
a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\) = \(\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\) = \(\dfrac{x+1}{2x}\)
c) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\) = \(\dfrac{\left(8-x\right)\left[-\left(x+2\right)\right]}{\left(x+2\right)^2}\) =\(\dfrac{8-x}{x+2}\)
