Question

giải hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\\\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=1,5\end{matrix}\right.\)

Answer 1

Đặt \(\dfrac{1}{x+y-3}=a;\dfrac{1}{x-y+1}=b\)

Ta có hệ phương trình:

\(\left\{{}\begin{matrix}5a-2b=8\\3a+b=1.5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=1\\x-y+1=\dfrac{-3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\x-y=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{13}{4}\end{matrix}\right.\)