| x | -3 | 1 | |||
| x+3 | - | 0 | + | \(|\) | + |
| x-1 | - | \(|\) | - | 0 | + |
+) Nếu \(-3\le x\Leftrightarrow|x-1|=1-x\)
\(|x+3|=-x-3\)
\(pt\Leftrightarrow1-x-x-3=5\)
\(\Leftrightarrow-2x-2=5\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\frac{-7}{2}\left(tm\right)\)
+) Nếu \(-3< x< 1\Leftrightarrow|x-1|=1-x\)
\(|x+3|=x+3\)
\(pt\Leftrightarrow1-x+x+3=5\)
\(\Leftrightarrow4=5\) ( vô lí )
+) Nếu \(x\ge1\Leftrightarrow|x-1|=x-1\)
\(|x+3|=x+3\)
\(pt\Leftrightarrow x-1+x+3=5\)
\(\Leftrightarrow2x+2=5\)
\(\Leftrightarrow x=\frac{3}{2}\left(tm\right)\)
Vậy ....
Ta có:\(|x-1|\ge0\)
\(|x+3|\ge0\)
Theo bài:
\(|x-1|+|x+3|=5\)
\(\rightarrow x-1+x+3=5\)
\(\rightarrow\left(x+x\right)+[\left(-1\right)+3]=5\)
\(\rightarrow2x+2=5\)
\(\rightarrow2x=5-2\)
\(\rightarrow2x=3\)
\(\rightarrow x=3:2\)
\(\rightarrow x=\frac{3}{2}\)
