\(ĐK:x\ge3;y\ne-1\)
Đặt \(\sqrt{x-3}=a;a\ge0\)
\(\dfrac{1}{y+1}=b\)
Khi đó, hpt trở thành:
\(\left\{{}\begin{matrix}3a-b=1\\a+2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a-2b=2\\a+2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7a=7\\a+2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\1+2b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}=1\\\dfrac{1}{y+1}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-3=1\\y+1=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-\dfrac{1}{2}\end{matrix}\right.\) ( tm )
Đặt a = \(\sqrt{x-3}\) và b = \(\dfrac{1}{y+1}\) ta có hệ phương trình
\(\left\{{}\begin{matrix}3a-b=1\\a+2b=5\end{matrix}\right.=>\left\{{}\begin{matrix}b=3a-1\\a+2\left(3a-1\right)=5\end{matrix}\right.=>\left\{{}\begin{matrix}b=3a-1\\a+6a-2=5\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}b=3a-1\\7a=7\end{matrix}\right.=>\left\{{}\begin{matrix}b=3a-1\\a=1\end{matrix}\right.=>\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\)
Thay a và b ta có
\(\left\{{}\begin{matrix}\sqrt{x-3}=1\\\dfrac{1}{y+1}=2\end{matrix}\right.=>\left\{{}\begin{matrix}x-3=1\\2\left(y+1\right)=1\end{matrix}\right.=>\left\{{}\begin{matrix}x=4\\2y+2=1\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}x=4\\y=\dfrac{-1}{2}\end{matrix}\right.\) Vậy hpt có nghiệm duy nhất (x;y)=(4;-1/2)
