Question

giải phương trình sau:

3|x - 1|+2|x - 4| = x+17

Answer 1

TH1: \(x\le1\) <=> \(\left\{{}\begin{matrix}\left|x-1\right|=1-x\\\left|x-4\right|=4-x\end{matrix}\right.\)

PT <=> \(3\left(1-x\right)+2\left(4-x\right)=x+17\)

<=> x = -1 (chọn)

TH2: \(1< x< 4< =>\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-4\right|=4-x\end{matrix}\right.\)

PT <=> 3(x-1) + 2(4-x) = x+17

<=> x = \(\varnothing\)

TH3: \(x\ge4< =>\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-4\right|=x-4\end{matrix}\right.\)

PT <=> 3(x-1) + 2(x-4) = x+17

<=> x = 7 (chọn)

KL: \(S=\left\{-1;7\right\}\)