\(\hept{\begin{cases}xy+x+y=x^2-2y^2\left(1\right)\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\left(2\right)\end{cases}}\)

Đk:\(x\ge1;y\ge0\)

\(\left(1\right)\Leftrightarrow2y^2\left(x+1\right)y-x^2+x=0\)

\(\Delta=\left(x+1\right)^2-8\left(-x^2+x\right)=x^2+2x+1+8x^2-8x\)

\(=9x^2-6x+1=\left(3x-1\right)^2\)

Do \(x\ge1\Rightarrow\sqrt{\Delta}=3x-1>0\)

\(\Rightarrow\orbr{\begin{cases}y_1=\frac{-x-1-\left(3x-1\right)}{4}=-x\\y_2=\frac{-x-1+\left(3x-1\right)}{4}=\frac{x-1}{2}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-y\\x=2y+1\end{cases}}\)

\(\left(2y+1\right)\sqrt{2y}-y\sqrt{2y}=4y+2-2y\)

\(\Leftrightarrow y\sqrt{2y}+\sqrt{2y}-2y-2=0\)

\(\Leftrightarrow\sqrt{2}\left(\sqrt{y}\right)^3-2\left(\sqrt{y}\right)^2+\sqrt{2}\left(\sqrt{y}\right)-2=0\)

\(\Leftrightarrow\sqrt{y}=\sqrt{2}\Leftrightarrow y=2\Rightarrow x=5\)

Vậy...

1+1+1+1=4

k mk mk k lại

Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)

\(x.y=10\Rightarrow2k.5k=10\Rightarrow10k^2=10\Rightarrow k^2=1\Rightarrow k=1\)

\(\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}\)