Question

[Đặng Thị Hiền Minh]

Rút gọn:

[GTTĐ ( x-1) + GTTĐ ( x )+x ] / ( 3x^2 - 4x +1), vs x < 0

Answer 1

[Thu Thao]

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

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 :))))