\(ĐK:x\le6;y\ge3\\ \left\{{}\begin{matrix}x^2+2y=xy+4\left(1\right)\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-4+2y-xy=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)-y\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-y+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=y-2\end{matrix}\right.\)
Từ đó thế vào PT(2)
Với \(x=y-2\Leftrightarrow x+2=y\)
\(\left(2\right)\Leftrightarrow x^2-x+3-x\sqrt{6-x}=\left(x-1\right)\sqrt{x-1}\left(1\le x\le6\right)\\ \Leftrightarrow2x^2-2x+6-2x\sqrt{6-x}=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+x\left(x-1\right)=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+\left(x-1\right)\left(x-2\sqrt{x-1}\right)=0\\ \Leftrightarrow\left(\dfrac{x^2-6+x}{x+\sqrt{6-x}}\right)^2+\dfrac{\left(x-1\right)\left(x^2-4x+4\right)}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left[\dfrac{\left(x-2\right)\left(x+3\right)}{x+\sqrt{6-x}}\right]^2+\dfrac{\left(x-1\right)\left(x-2\right)^2}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left(x-2\right)^2\left[\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}=0\left(1\right)\end{matrix}\right.\)
Dễ thấy \(\left(1\right)>0\) với \(x\ge1\)
Do đó \(x=2\Leftrightarrow y=4\)
Vậy HPT có nghiệm \(\left(x;y\right)=\left(2;4\right)\)
