Question

Giải hệ phương trình:\(\left\{{}\begin{matrix}2x+3y=xy+5\\\dfrac{1}{x}+\dfrac{1}{y+1}=1\end{matrix}\right.\)

Answer 1

Đk: \(x\ne0,y\ne-1\)

\(\left\{{}\begin{matrix}2x+3y=xy+5\left(1\right)\\\dfrac{1}{x}+\dfrac{1}{y+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1+x=x\left(y+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1=xy\end{matrix}\right.\)

\(\Rightarrow2x+3y=y+1+5\)

\(\Leftrightarrow x=3-y\) thay vào (1) có:

\(2\left(3-y\right)+3y=\left(3-y\right)y+5\)

\(\Leftrightarrow y^2-2y+1=0\)

\(\Leftrightarrow y=1\) \(\Rightarrow x=2\)(tm)

Vậy (x;y)=(2;1)