\(\left|x+1\right|+\left|1-x\right|=2\)
\(\Leftrightarrow\left|x+1\right|+\left|x-1\right|=2\)
\(\Leftrightarrow\left|x+1\right|+\left|x+1-1\right|=2\)
\(\Leftrightarrow\)\(\left|x+1\right|+\left|x+1\right|-\left|1\right|=2\)
\(\Leftrightarrow2\left|x+1\right|-1=2\)
\(\Leftrightarrow2\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(x+1\right)=3khi:x+1\ge0\\2\left(-x-1\right)=3khi:x+1< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}khi:x\ge1\left(loại\right)\\x=-\frac{5}{2}khi:x< 1\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x =\(-\frac{5}{2}\)
+ \(\left|x+1\right|\ge x+1\forall x\). Dấu "=" xảy ra \(\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)
\(\left|1-x\right|\ge1-x\forall x\). Dấu "=" xảy ra \(\Leftrightarrow1-x\ge0\Leftrightarrow x\le1\)
Do đó : \(\left|x+1\right|+\left|1-x\right|\ge x+1+1-x\)
\(\Rightarrow\left|x+1\right|+\left|1-x\right|\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\Leftrightarrow-1\le x\le1\)
