Question

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

Answer 1

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

Lấy \(\left(1\right)-\left(2\right)\)

\(\Rightarrow\left(\dfrac{3}{x}-\dfrac{1}{x}\right)+\left(\dfrac{1}{y}-\dfrac{1}{y}\right)=2-1\)

\(\Rightarrow\dfrac{2}{x}=1\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào \(\left(1\right):\)

\(\Rightarrow\dfrac{3}{2}+\dfrac{1}{y}=2\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow y=2\)

Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(2;2\right)\)