a) ta xét các trường hợp:
+ Với x \(\)<-1
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4-x+3+x+1=5\)
\(\Rightarrow-x+8=5\)
\(\Rightarrow-x=-3\)
\(\Rightarrow x=3\)(không thỏa mãn )
+Với -1\(\le\)x<3
\(\)\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4-x+3-x-1=5\)
\(\Rightarrow-3x+6=5\)
\(\Rightarrow-3x=-1\)
\(\Rightarrow x=\frac{1}{3}\)(thỏa mãn)
+ Với 3\(\le\)x<4
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4+x-3-x-1=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)(không thỏa mãn)
+ Với x\(\ge\)4
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow x-4+x-3-x-1=5\)
\(\Rightarrow x-8=5\)
\(\Rightarrow x=13\)(thỏa mãn)
Vậy \(x\in\left\{\frac{1}{3};13\right\}\)thì \(\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
