Đáp án:
Đặt `1/(x+3)=a` (`x \ne -3`)
`1/(y-5)=b` (`y \ne 5`)
Ta có :
$\left\{\begin{matrix}a+2b=4\\ 3a-4b=3\end{matrix}\right.$
$⇔\left\{\begin{matrix}2a+4b=8\\ 3a-4b=3\end{matrix}\right.$
$⇔\left\{\begin{matrix}5a=11\\ a+2b=4\end{matrix}\right.$
$⇔\left\{\begin{matrix}a=\dfrac{11}{5}\\ b=\dfrac{9}{10}\end{matrix}\right.$
$⇔\left\{\begin{matrix}\dfrac{1}{x+3}=\dfrac{11}{5}\\ \dfrac{1}{y-5}=\dfrac{9}{10}\end{matrix}\right.$
$⇔\left\{\begin{matrix}x+3=\dfrac{5}{11}\\ y-5=\dfrac{10}{9}\end{matrix}\right.$
$⇔\left\{\begin{matrix}x=\dfrac{-28}{11}\\ y=\dfrac{55}{9}\end{matrix}\right.$
Vậy hệ phương trình có nghiệm duy nhất `(x;y)=(-28/11;55/9)`
