\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+4\sqrt{xy}=16\\x+y=10\end{matrix}\right.\) ( ĐK : \(x,y\ge0\) )
Đặt : \(\sqrt{x}+\sqrt{y}=a\) ; \(\sqrt{xy}=b\) ; \(a,b\ge0\) Phương tình trở thành :
\(\left\{{}\begin{matrix}a+4b=16\\a^2-2b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+4b=16\\2a^2-4b=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+2a^2=36\\a+4b=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a^2+a-36=0\\a+4b=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=4\\\sqrt{xy}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\xy=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10-x\\x\left(10-x\right)=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10-x\\-x^2+10x-9=0\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}x=9\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=9\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;9\right)\) hoặc \(\left(x;y\right)=\left(9;1\right)\)
điều kiện : \(0\le x;y\le10\)
ta có : \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+4\sqrt{xy}=16\\x+y=10\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\sqrt{y}+2\left(\sqrt{x}+\sqrt{y}\right)^2=36\)
\(\Leftrightarrow2\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}+\sqrt{y}\right)-36=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{y}=4\\\sqrt{x}+\sqrt{y}=\dfrac{-9}{2}\left(loại\right)\end{matrix}\right.\)
từ \(\sqrt{x}+\sqrt{y}=4\Rightarrow4+4\sqrt{xy}=16\Leftrightarrow\sqrt{xy}=3\)
\(\Rightarrow\) biến đổi nảy giờ ta có được : \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=4\\\sqrt{xy}=3\end{matrix}\right.\)
\(\Rightarrow\) \(\sqrt{x};\sqrt{y}\) là 2 nghiệm của phương trình : \(X^2-4X+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y}=3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=9\end{matrix}\right.\end{matrix}\right.\) vậy phương trình có 2 nghiệm \(\left(9;1\right);\left(1;9\right)\)