Question

Giúp em bài này với ạ!!!!

 

Answer 1

Đặt  \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\left(a;b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}a-2b=2.\\5a-4b=5.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}.\\b=\dfrac{-5}{6}.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=\dfrac{1}{3}.\\\dfrac{1}{x-y}=\dfrac{-5}{6}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=3.\\-5x+5y=6.\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{10}.\\y=\dfrac{21}{10}.\end{matrix}\right.\)