đặt \(\sqrt{x-2}\)=a \(\sqrt{y-3}\)=b
2a+3b=14
\(\left\{{}\begin{matrix}2a+3b=14\\a+b=5\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-b\\2.\left(5-b\right)+3b=14\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=5-b\\10-2b+3b=14\end{matrix}\right.\)\(\left\{{}\begin{matrix}a=5-b\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-4\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=1\\b=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{1}\\\sqrt{y-3}=\sqrt{16}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2=1\\y-3=16\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=3\\y=19\end{matrix}\right.\)
⇒phương trình có 2 no (x,y)=(3, 19)
