a) Áp dụng bđt |a| + |b| \(\ge\) |a+b| ta có:
\(\left|x-1\right|+\left|x+3\right|=\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|\)
\(\ge\left|4\right|=4\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}x-1\le0\\x+3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\le1\\x\ge-3\end{matrix}\right.\)\(\Leftrightarrow-3\le x\le1\)
b) Xét từng khoảng
+ \(x< -\frac{3}{2}\)
+ \(-\frac{3}{2}\le x< 4\)
+ \(x\ge4\)
a) Vì \(\left|x-1\right|+\left|x+3\right|=4\)
\(\Rightarrow\left|1-x\right|+\left|x+3\right|=4\)
Nhận thấy \(\left[{}\begin{matrix}\left|1-x\right|\ge1-x\forall x\\\left|x+3\right|\ge x+3\forall x\end{matrix}\right.\)
\(\Rightarrow\left|1-x\right|+\left|x+3\right|\ge1-x+x+3\)
\(\Rightarrow\left|1-x\right|+\left|x+3\right|\ge4\)
Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}1-x\ge0\\x+3\ge0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-3\end{matrix}\right.\) \(\Rightarrow-3\le x\le1\)
\(\Rightarrow x\in\left\{-3-2;-1;0;1\right\}\)
Vậy \(x\in\left\{-3;-2;-1;0;1\right\}\).