Ta có:
\(x\left(x-y\right)=\dfrac{3}{10}\)
\(x^2-xy=\dfrac{3}{10}\) (1)
\(y\left(x-y\right)=\dfrac{-3}{50}\)
\(xy-y^2=\dfrac{-3}{50}\) (2)
Trừ vế với vế của (1) cho (2), ta có:
\(x^2-xy-xy+y^2=\dfrac{3}{10}-\dfrac{-3}{50}\)
\(\left(x-y\right)^2=\dfrac{18}{50}=\dfrac{9}{25}\)
\(x-y=\dfrac{3}{5}\) hoặc \(x-y=\dfrac{-3}{5}\)
*) \(x-y=\dfrac{3}{5}\)
\(x\left(x-y\right)=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}:\left(x-y\right)\)
\(x=\dfrac{3}{10}:\dfrac{3}{5}\)
\(x=\dfrac{1}{2}\)
\(y=\dfrac{1}{2}-\dfrac{3}{5}\)
\(y=-\dfrac{1}{10}\)
*) \(x-y=\dfrac{-3}{5}\)
\(x\left(x-y\right)=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}:\left(x-y\right)\)
\(x=\dfrac{3}{10}:\dfrac{-3}{5}\)
\(x=\dfrac{-1}{2}\)
\(y=\dfrac{-1}{2}-\dfrac{-3}{5}\)
\(y=\dfrac{1}{10}\)
Vậy ta được các cặp giá trị (x; y) sau:
\(\left(\dfrac{1}{2};\dfrac{-1}{10}\right);\left(\dfrac{-1}{2};\dfrac{1}{10}\right)\)
