Question

\(\left\{{}\begin{matrix}2y=\dfrac{y+1}{x}\\2x=\dfrac{x+1}{y}\end{matrix}\right.\)

Answer 1

\(ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2xy=y+1\\2xy=x+1\end{matrix}\right.\Leftrightarrow y+1=x+1\Leftrightarrow x=y\)

Thế vào PT (1) ta được \(2x=\dfrac{x+1}{x}\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy HPT có nghiệm \(\left(x;y\right)\) là \(\left(1;1\right);\left(-\dfrac{1}{2};-\dfrac{1}{2}\right)\)