\(\left\{{}\begin{matrix}3\sqrt{2x-1}-\dfrac{y}{y+1}=1\\\sqrt{2x-1}+\dfrac{2y}{y+1}=5\end{matrix}\right.\left(x\ge\dfrac{1}{2};y\ne-1\right)\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x-1}\\b=\dfrac{y}{y+1}\end{matrix}\right.\left(a\ge0\right)\)
hệ pt trở thành \(\left\{{}\begin{matrix}3a-b=1\\a+2b=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6a-2b=2\left(1\right)\\a+2b=5\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right)\Rightarrow7a=7\Rightarrow a=1\Rightarrow b=2\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{2x-1}=1\\\dfrac{y}{y+1}=2\end{matrix}\right.\)
\(\sqrt{2x-1}=1\Rightarrow x=1\)
\(\dfrac{y}{y+1}=2\Rightarrow2y+2=y\Rightarrow y=-2\)
Vậy hệ có bộ nghiệm (x,y) là \(\left(1,-2\right)\)
