Question

giải hệ phương trình

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

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\y\ge0\end{matrix}\right.\) \(\Rightarrow x+y>0\)

\(xy+x+y=x^2-2y^2\Leftrightarrow\left(x^2-xy-2y^2\right)-\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)-\left(x+y\right)=0\)

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

\(\Leftrightarrow x-2y-1=0\Rightarrow x=2y+1\)

Thay xuống pt dưới:

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

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

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