Question

M=\(\dfrac{|x+1|+2x}{3x^2-2x-1}\)

Rút gọn M

Answer 1

Ta có: \(3x^2-2x-1=3x^2-3x+x-1\)

\(=3x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(3x+1\right)\)

Ta có: \(\left|x+1\right|=x+1\) nếu \(x+1\ge0\Rightarrow x\ge-1\)

\(\left|x+1\right|=-x-1\) nếu \(x+1\le0\Rightarrow x\le-1\)

*Với \(x\ge-1\), ta có:

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

*Với \(x\le-1\) ta có:

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

Vậy..........

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