Đk: \(\left\{{}\begin{matrix}y\ge0\\x>1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{9\left(x-1\right)y}=y\left(2+\sqrt{\dfrac{y}{x-1}}\right)\left(1\right)\\y^2+xy-5x+7=0\left(2\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{\left(x-1\right)y}\left(a\ge0\right)\\b=\sqrt{\dfrac{y}{x-1}}\left(b\ge0\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3a=ab\left(2+b\right)\)
Với \(a=0\Rightarrow\sqrt{\left(x-1\right)y}=0\Rightarrow y=0\) (vì \(x\ne1\)).
Thay \(y=0\) vào (2) ta được:
\(2^2+x.2-5x+7=0\)
\(\Leftrightarrow x=\dfrac{11}{3}\left(nhận\right)\)
Với \(a\ne0\Rightarrow3=b\left(2+b\right)\)
\(\Leftrightarrow b^2+2b-3=0\)
\(\Leftrightarrow b^2-b+3b-3=0\)
\(\Leftrightarrow b\left(b-1\right)+3\left(b-1\right)=0\)
\(\Leftrightarrow\left(b-1\right)\left(b+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\left(nhận\right)\\b=-3\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{y}{x-1}}=1\Rightarrow x=y+1\)
Thay vào (2) ta được:
\(y^2+\left(y+1\right)y-5\left(y+1\right)+7=0\)
\(\Leftrightarrow y^2+y^2+y-5y-5+7=0\)
\(\Leftrightarrow2y^2-4y+2=0\)
\(\Leftrightarrow2\left(y-1\right)^2=0\)
\(\Leftrightarrow y=1\Rightarrow x=1+1=2\)
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{11}{3};0\right),\left(2;1\right)\right\}\)
