\(\left\{\begin{matrix} x^{2}+xy-2y^{2}+3y-1=\sqrt{y-1}-\sqrt{x}\; (1)\\3\sqrt{6-y}+3\sqrt{2x+3y-7}=2x+7\; \; \; (2) \end{matrix}\right.\)
ĐK \(\left\{\begin{matrix} x\geq 0\\1\leq y\leq 6 \\2x+3y-7\geq 0 \end{matrix}\right.(*)\)
Từ (1) \(\Rightarrow \sqrt{y-1}-\sqrt{x}+(y-1)^{2}-x^{2}+y(y-x-1)=0\)
\(\Leftrightarrow (y-x-1)\left ( \underset{>0,x\geq 0 \& 6\geq y\geq 1}{\underbrace{\frac{1}{\sqrt{y-1}+x}+2y+x-1}} \right )=0\Rightarrow y-x-1=0\Leftrightarrow x=y-1\; \; (3)\)
Thế (3) vào (2) ta được pt \(3\sqrt{6-y}+3\sqrt{5y-9}=2y+5,\; (4)\) đk \(\frac{9}{5}\leq y \leq 6\)
Giải (4) \(\Leftrightarrow (8-y)-3\sqrt{6-y}+3(y-1-\sqrt{5y-9})=0\)
\(\Leftrightarrow \frac{y^{2}-7y+10}{(8-y)+3\sqrt{6-y}}+3.\frac{y^{2}-7y+10}{y-1+\sqrt{5y-9}}=0\)
\(\Leftrightarrow (y^{2}-7y+10)(\underset{>0,\forall \frac{9}{5}\leq y\leq 6}{\underbrace{{\frac{1}{(8-y)+3\sqrt{6-y}}+\frac{3}{y-1+\sqrt{5y-9}}}}})=0\)
\(\Leftrightarrow y^{2}-7y+10=0\Leftrightarrow \Bigg \lbrack\begin{matrix} y=2\underset{(4)}{\underbrace{\rightarrow}} x=1(tm(*))\\y=5\underset{(4)}{\underbrace{\rightarrow}} x=4(tm(*)) \end{matrix}\)
Vậy hpt có hai nghiệm \((x;y)=(1;2),(x;y)=(4;5)\)
