Question

\(\left|x-1\right|-2\left|x-2\right|+3\left|x-3\right|=14\)

Answer 1

|x-1|=\(\left\{{}\begin{matrix}x-1neux-1\ge0\Leftrightarrow x\ge1\\-\left(x-1\right)neux-1< 0\Leftrightarrow x< 1\end{matrix}\right.\)

|x-2|=\(\left\{{}\begin{matrix}x-2neux-2\ge0\Leftrightarrow x\ge2\\-\left(x-2\right)neux-2< 0\Leftrightarrow x< 2\end{matrix}\right.\) |x-3|=\(\left\{{}\begin{matrix}x-3neux-3\ge0\Leftrightarrow x\ge3\\-\left(x-3\right)neux-3< 0\Leftrightarrow x< 3\end{matrix}\right.\) TH1:x≥1;2;3 |x-1|=x-1;|x-2|=x-2;|x-3|=x-3 x-1-2.(x-2)+3.(x-3)=14 ⇔x-1-2x+4+3x-9=14 ⇔x-2x+3x=14+1-4+9 ⇔2x=20⇔x=10(t/m) TH2:x<1;2;3 |x-1|=-(x-1)=-x+1 |x-2|=-(x-2)=-x+2 |x-3|=-(x-3)=-x+3 -x+1-2.(-x+2)+3.(-x+3)=14 ⇔-x+1+2x-4-3x+9=14 ⇔-x+2x-3x=14-1+4-9 ⇔-2x=8⇔x=-4(t/m) vậy PT có tập nghiệm S=\(\left\{-4;10\right\}\)