Question

Giải hệ phương trình \(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=11\\\sqrt{x+3}+2\left|y-2\right|=10\end{matrix}\right.\)

Answer 1

ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{x+3}+2\left|y-2\right|=22\\\sqrt{x+3}+2\left|y-2\right|=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+3}=4\\\left|y-2\right|=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=16\\\left[{}\begin{matrix}y-2=3\\y-2=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=13\\\left[{}\begin{matrix}y=5\\y=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(13;5\right);\left(13;-1\right)\)