Question

Giải hệ phương trình \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y-1}=1\\3y-1=xy\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y-1}=1\left(1\right)\\3x-1=xy\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{x}+\frac{1}{y-1}=1\)

\(\Leftrightarrow\frac{y-1+x}{x\left(y-1\right)}=1\)

\(\Leftrightarrow x+y-1=xy-x\)

\(\Leftrightarrow2x+y-1=xy\)

Lại theo (2) nên ta có :

\(2x+y-1=3y-1\)

\(\Leftrightarrow2x-2y=0\)

\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

Thay lên ta được : \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{x-1}=1\\3y-1=y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3\pm\sqrt{5}}{2}\\y=\frac{3\pm\sqrt{5}}{2}\end{matrix}\right.\)