Bài 1:
a) \(\dfrac{\sqrt{x^2+2x+1}}{\left|x\right|-1}=\dfrac{\sqrt{\left(x+1\right)^2}}{\left|x-1\right|}=\dfrac{x+1}{\left|x\right|-1}\)
Bài 2:
a) \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=0\) (1)
\(\Leftrightarrow\sqrt{x^2-2x+1}=-\sqrt{x^2-4x+4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x^2-2x+1}=0\\-\sqrt{x^2-4x+4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\Leftrightarrow x\in\varnothing\)
Vậy phương trình (1) vô nghiệm.