Question

Rút gọn rồi tính GTBT:

\(A=\dfrac{\left(x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\) với \(x=\dfrac{1}{2}\)

Answer 1

\(A=\dfrac{\left(x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{x\left(x+2\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)

Với \(x=\dfrac{1}{2}\) ta có: \(A=\dfrac{\dfrac{1}{2}-2}{\dfrac{1}{2}+1}=\dfrac{-\dfrac{3}{2}}{\dfrac{3}{2}}=-1\)

Answer 2

\(A=\dfrac{\left(x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+2\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+2\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

\(=\dfrac{x-2}{x+1}\)

Thay \(x=\dfrac{1}{2}\) vào bt ta dc :

\(A=\dfrac{\dfrac{1}{2}-2}{\dfrac{1}{2}+1}=-1\)

Vậy...