Question

[Cỏ dại]

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

[Nguyệt]

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

\(A=\frac{\left(x^2+2x\right).\left(x^2-4x+4\right)}{\left(x^3-4x\right).\left(x+1\right)}=\frac{x^4-4x^3+4x^2+2x^3-8x^2+8x}{x^4+x^3-4x^2-4x}\)

\(A=\frac{x^4-2x^3-4x^2+8x}{x^4+x^3-4x^2-4x}=\frac{x^3.\left(x-2\right)-4x.\left(x-2\right)}{x^3.\left(x+1\right)-4x.\left(x+1\right)}=\frac{\left(x^3-4x\right).\left(x-2\right)}{\left(x^3-4x\right).\left(x+1\right)}=\frac{x-2}{x+1}\)

thay \(x=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}-2}{\frac{1}{2}+1}=\frac{-\frac{3}{2}}{\frac{3}{2}}=-1\)

Vậy A=-1