- Đặt \(\left|x-1\right|=a,\left|y+2\right|=b\) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}3a-2b=4\\2a+3b=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}6a-4b=8\\6a+9b=21\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3a-2b=4\\-13b=-13\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3a-2=4\\b=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) ( I )
- Thay \(\left|x-1\right|=a,\left|y+2\right|=b\) vào ( I ) ta được :
- Ta có : \(\left|x-1\right|=2\)
TH1 : \(x-1\ge0\left(x\ge1\right)\)
=> \(\left|x-1\right|=x-1=2\)
=> \(x=3\left(TM\right)\)
TH2 : \(x-1< 0\left(x< 1\right)\)
=> \(\left|x-1\right|=1-x=2\)
=> \(x=-1\left(TM\right)\)
Ta có : \(\left|y+2\right|=1\)
TH1 : \(y+2\ge0\left(y\ge-2\right)\)
=> \(\left|y+2\right|=y+2=1\)
=> \(y=-1\left(TM\right)\)
TH2 : \(y+2< 0\left(y< -2\right)\)
=> \(\left|y+2\right|=-y-2=1\)
=> \(y=-3\left(TM\right)\)
Vậy hệ phương trình có các cặp nghiệm là \(\left(x,y\right)=\left(3,-3\right),\left(x,y\right)=\left(-1,-1\right),\left(x,y\right)=\left(3,-1\right),\left(x,y\right)=\left(-1,-3\right)\)
\(\left\{{}\begin{matrix}3\left|x-1\right|-2\left|y+2\right|=4\\2\left|x-1\right|+3\left|y+2\right|=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left|x-1\right|-4\left|y+2\right|=8\\6\left|x-1\right|+9\left|y+2\right|=21\end{matrix}\right.\)
\(\Rightarrow-13\left|y+2\right|=-13\Leftrightarrow\left|y+2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=-3\end{matrix}\right.\)
Thay vô tính x là xong
