TXĐ: \(x\ne1\)
\(\dfrac{\sqrt{x^2-2x+1}}{1-x}=\dfrac{\sqrt{\left(x-1\right)^2}}{1-x}=-\dfrac{\left|x-1\right|}{x-1}\)
TH1: \(x-1>0,\) ta có:
\(-\dfrac{\left|x-1\right|}{x-1}=\)\(-\dfrac{x-1}{x-1}=-1\)
TH2:\(x-1< 0\)
\(-\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
a. \(\dfrac{\sqrt{x^2-2x+1}}{1-x}=\dfrac{\sqrt{\left(x-1\right)^2}}{1-x}=\dfrac{\sqrt{\left(1-x\right)^2}}{1-x}=\dfrac{1-x}{1-x}=1\)
a)+)ĐKXĐ:\(\left\{{}\begin{matrix}x^2-2x+1\\1-x\ne0\end{matrix}\right.\)
=>\(\left(x-1\right)^2>0\)
=> \(x-1>0\)
=>\(x>1\)
+)Tính:
\(\dfrac{\sqrt{x^2-2x+1}}{1-x}=\dfrac{\sqrt{\left(x-1\right)^2}}{1-x}=\dfrac{\sqrt{\left(1-x\right)^2}}{1-x}=\dfrac{1-x}{1-x}=1\)
