b) TH1 :\(x< \frac{1}{2};\)ta có :
\(3-6x+x-5=7-2x\)
\(-2-3x=7\)
\(3x=-9\)
\(x=-3\left(TM\right)\)
TH2 : \(x\ge\frac{1}{2};\)ta có :
\(6x-3+x-5=7-2x\)
\(9x-8=7\)
\(9x=15\)
\(x=\frac{5}{3}\left(TM\right)\)
Vậy ...
a) Do \(\left|x+y-3\right|+\left|x+1\right|=4\)
\(\Rightarrow\left|x+1\right|\le4\)
\(\Rightarrow\hept{\begin{cases}x+1\le4\\x+1\ge-4\end{cases}}\)
\(\Rightarrow-3\le x\le3\)
Thử chọn \(x\)để tìm y tương ứng.