\(\frac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\)
Có
x < 0
=> x - 1 < 0
=> | x - 1 | = 1 - x
Khi đó \(\left|x-1\right|+\left|x\right|+x=1-x-x+x=1-x\)
Mặt khác ta có
\(3x^2-4x+1=\left(3x^2-3x\right)-\left(x-1\right)=3x\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(3x-1\right)\)Do đó
\(\frac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}=\frac{1-x}{\left(x-1\right)\left(3x-1\right)}=-\frac{1}{3x-1}\)
Ko chắc lém :))))