Ta có: \(\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
* Xét \(x\ge1\Rightarrow\left|x+1\right|=x+1\Rightarrow Q=2x-\left(x+1\right)=x-1\)* Xét \(x< -1\Rightarrow\left|x+1\right|=-x-1\Rightarrow Q=2x-\left(-x-1\right)=3x+1\)Vậy \(Q=x-1\) khi \(x\ge-1\) và \(Q=3x+1\) khi \(x< -1\)