Question

Rút gọn biểu thức : \(A=\dfrac{|x-2|+|x|+x}{3x^2-8x+4}\)

Answer 1

Trường hợp 1: x<0

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

Trường hợp 2: 0<=x<2

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

TRường hợp 3: x>=2

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