Question

\(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\)

giải hpt

Answer 1

ĐKXĐ: x>=0 và y>=1

\(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\8\sqrt{x}-2\sqrt{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}=9\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=4-2=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y-1=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\left(nhận\right)\)