Question

giải \(\left\{{}\begin{matrix}x+y+\dfrac{1}{y}=\dfrac{9}{x}\\x+y-\dfrac{4}{x}=\dfrac{4y}{x^2}\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}x+y+\dfrac{1}{y}=\dfrac{9}{x}\left(1\right)\\x+y-\dfrac{4}{x}=\dfrac{4y}{x^2}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)-\left(2\right)\) vế theo vế

\(\Rightarrow\dfrac{1}{y}+\dfrac{4}{x}=\dfrac{9}{x}-\dfrac{4y}{x^2}\)

\(\Leftrightarrow\dfrac{4y}{x^2}+\dfrac{1}{y}-\dfrac{5}{x}=0\)

\(\Rightarrow4y^2+x^2-5xy=0\)

\(\Leftrightarrow\left(y-x\right)\left(4y-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=4y\end{matrix}\right.\)

Xét từng trường hợp rồi rút ra kết luận nhé (^ . ^)