Question

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

Answer 1

ĐKXĐ: x<>-2 và y>=3

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

=>\(\left\{{}\begin{matrix}\dfrac{2}{x+2}+4\sqrt{y-3}=14\\\dfrac{2}{x+2}-3\sqrt{y-3}=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{y-3}=21\\\dfrac{2}{x+2}-3\sqrt{y-3}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{y-3}=3\\\dfrac{2}{x+2}=-7+3\cdot3=9-7=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y-3=9\\x+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=-1\end{matrix}\right.\left(nhận\right)\)