Question

Giải hệ pt:

\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y+2\right|=2\\4\left|x-1\right|+3\left|y+2\right|=7\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y+2\right|=2\\4\left|x-1\right|+3\left|y+2\right|=7\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y+2\right|=2\\4x-4+3y+6=7\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y=1\\4x+3y=8\end{matrix}\right.\)

\(\left\{{}\begin{matrix}y=-4\\4x+3.-4=8\end{matrix}\right.\)

\(\left\{{}\begin{matrix}y=4\\x=5\end{matrix}\right.\)

Vậy HPT có 1 nghiệm duy nhất (5;4)

Answer 2

Đặt |x-1|=a; |y+2|=b

Theo đề, ta có; a+b=2 và 4a+3b=7

=>a=1; b=1

=>|x-1|=1 và |y+2|=1

=>\(\left\{{}\begin{matrix}x\in\left\{2;0\right\}\\y\in\left\{-1;-3\right\}\end{matrix}\right.\)