a, \(\left\{{}\begin{matrix}2\left(\sqrt{5}+2\right)x+2y=6-2\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\)
-> hpt vô nghiệm
c, đặt 1/x = a ; 1/y = b
\(\left\{{}\begin{matrix}2a-2b=2\\2a+3b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{5}\\a=\dfrac{8}{5}\end{matrix}\right.\)
Theo cách đặt x = 5/8 ; y = 5/3