Question

giải hệ phương trình \(\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x}+\dfrac{3}{y}=2\end{matrix}\right.\)

 

Answer 1

ĐKXĐ: \(x;y\ne0\)

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

\(\Leftrightarrow2\left(x+1\right)+3x=2x\left(x+1\right)\)

\(\Leftrightarrow2x^2-3x-2=0\Rightarrow\left[{}\begin{matrix}x=2\Rightarrow y=3\\x=-\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}\end{matrix}\right.\)