Question

Gỉải hệ pt:
\(\left\{{}\begin{matrix}\frac{108}{x+y}+\frac{63}{x-y}=7\\\frac{81}{x+y}+\frac{84}{x-y}=7\end{matrix}\right.\)

Answer 1

Đặt: \(\left\{{}\begin{matrix}\frac{1}{x+y}=a\\\frac{1}{x-y}=b\end{matrix}\right.\)

Hệ đã cho trở thành: \(\left\{{}\begin{matrix}108b+63a=7\\81b+84a=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{27}\\b=\frac{1}{21}\end{matrix}\right.\)

\(\Rightarrow\frac{1}{x+y}=\frac{1}{27}\Rightarrow x+y=27\)

Và: \(\frac{1}{x-y}=\frac{1}{21}\Rightarrow x-y=21\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=27\\x-y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=24\\y=3\end{matrix}\right.\)

Vậy ................