\(\left|x+2\right|-\left|x-1\right|< \dfrac{x-3}{2}\)
Lập bảng xét dấu:
+) Với \(x< -2\Leftrightarrow-\left(x+2\right)-\left(1-x\right)< \dfrac{x-3}{2}\)
\(\Leftrightarrow-x-2-1+x< \dfrac{x-3}{2}\\ \Leftrightarrow-3< \dfrac{x-3}{2}\\ \Leftrightarrow\dfrac{x-3}{2}>-3\\ \Leftrightarrow x-3>-6\\ \Leftrightarrow x>-3\\ \Leftrightarrow-3< x< -2\)
+) Với \(-2\le x< 1\Leftrightarrow\left(x+2\right)-\left(1-x\right)< \dfrac{x-3}{2}\)
\(\Leftrightarrow x+2-1+x< \dfrac{x-3}{2}\\ \Leftrightarrow2x+1< \dfrac{x-3}{2}\\ \Leftrightarrow4x+2< x-3\\ \Leftrightarrow4x-x< -3-2\\ \Leftrightarrow3x< -5\\ \Leftrightarrow x< -\dfrac{5}{3}\\ \Leftrightarrow-2\le x< -\dfrac{5}{3}\)
+) Với \(x\ge1\Leftrightarrow x+2-x+1< \dfrac{x-3}{2}\)
\(\Leftrightarrow3< \dfrac{x-3}{2}\\ \Leftrightarrow\dfrac{x-3}{2}>3\\ \Leftrightarrow x-3>6\\ \Leftrightarrow x>9\left(T/m\right)\\ \)
Vậy bất phương trình có tập nghiệm \(S=\left\{x|-3< x< -\dfrac{5}{3};x>9\right\}\)
