\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-5y=3\\2\left|x-1\right|+4y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-9y=15\\\left|x-1\right|+2y=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{15}{9}\\\left|x-1\right|+2\cdot\left(-\dfrac{15}{9}\right)=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{15}{9}\\\left|x-1\right|=-\dfrac{8}{3}\end{matrix}\right.\)(vô lí)
Vậy HPT vô nghiệm
\(\left\{{}\begin{matrix}2\left|x-1\right|-5y=3\\\left|x-1\right|+2y=-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-5y=3\\2\left|x-1\right|+4y=-12\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-9y=3-\left(-12\right)=15\\2\left|x-1\right|-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{5}{3}\\2\left|x-1\right|=5y+3=-\dfrac{25}{3}+3=-\dfrac{16}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{5}{3}\\\left|x-1\right|=-\dfrac{8}{3}\left(vôlý\right)\end{matrix}\right.\)
=>\(\left(x;y\right)\in\varnothing\)
