Đặt : \(\dfrac{1}{x+y}=u,\dfrac{1}{x-y}=v\) (1)
Hệ phương trình (I) trở thành:
\(\left\{{}\begin{matrix}u-v=2\\5u-4v=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5u-5v=10\\5u-4v=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-v=7\\u-v=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=-7\\u=-5\end{matrix}\right.\)
Thay u = - 5 , v = - 7 vào (1) , ta có:
\(\dfrac{1}{x+y}=-5\Rightarrow x+y=-\dfrac{1}{5}\) (*)
\(\dfrac{1}{x-y}=-7\Rightarrow x-y=-\dfrac{1}{7}\)(**)
Từ: (*) và (**),ta có hệ:\(\left\{{}\begin{matrix}x+y=-\dfrac{1}{5}\\x-y=-\dfrac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-\dfrac{12}{35}\\x+y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{6}{35}\\y=-\dfrac{1}{35}\end{matrix}\right.\)
Vậy: Hệ có nghiệm duy nhất: (x;y) = (-6/35;-1/35)
=.= hk tốt!!
