\(\hept{\begin{cases}x+\sqrt{y}+y\sqrt{x}=30\\x\sqrt{x}+y\sqrt{y}=35\end{cases}}\)
\(\left(x\sqrt{y}+y\sqrt{x}=30\right)+\left(x\sqrt{x}+y\sqrt{y}=35\right)\)
\(\Rightarrow\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)=65\)
\(\left(x\sqrt{x}+y\sqrt{y}=35\right)-\left(x\sqrt{y}+y\sqrt{x}=30\right)\)
\(\Rightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)=5\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}+\sqrt{7}\\b=\sqrt{xy}\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a^2-2b\right)a=65\\\left(a^2-4b\right)a=5\end{cases}}}\)
\(\left[\left(a^22b\right)a=65\right]-\left[\left(a^2-4b\right)a=5\right]\)
\(\Rightarrow2ab=60\Rightarrow ab=30\Rightarrow a^3=125\)
\(\Rightarrow a=5;b=6\)
Vì a = 5 và b = 6
\(\Rightarrow\hept{\begin{cases}\sqrt{x}+\sqrt{y}=5\\\sqrt{xy}=6\end{cases}}\)\(x^2-5x+6=0\)
\(\Rightarrow\left(\sqrt{x};\sqrt{y}\right)\in\left\{\left(2;3\right);\left(3;2\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(9;4\right);\left(4;9\right)\right\}\)
ĐK: \(x,y\ge0\)
\(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=30\\x\sqrt{x}+y\sqrt{y}=35\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=30\\\left(\sqrt{x}+\sqrt{y}\right).\left[\left(\sqrt{x}+\sqrt{y}\right)^2-3.\sqrt{xy}\right]=35\end{cases}}\)
Đặt \(a=\sqrt{xy}\left(a\ge0\right),b=\sqrt{x}+\sqrt{y}\left(b\ge0\right)\), hệ trở thành :
\(\hept{\begin{cases}ab=30\\b.\left(b^2-3a\right)=35\end{cases}\Leftrightarrow}\hept{\begin{cases}ab=30\\b^3-3ab=35\end{cases}}\Leftrightarrow\hept{\begin{cases}ab=30\\b^3-3.30=35\end{cases}}\)
Từ đó tính ra b, rồi tính ra a, rồi tính ra x,y
