Question

\(\left\{{}\begin{matrix}6\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=1\\\\5\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{3}{x}=1\end{matrix}\right.\)

Answer 1

Gọi a=\(\dfrac{1}{x}\), b=\(\dfrac{1}{y}\)

⇔ \(\left\{{}\begin{matrix}6\left(a+b\right)=1\\5\left(a+b\right)+3a=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}6a+6b=1\\5a+5b+3a=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}6a+6b=1\\8a+5b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}24a+24b=4\\24a+15b=3\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}9b=1\\8a+5b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{9}\\8a+5.\dfrac{1}{9}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{9}\\a=\dfrac{1}{18}\end{matrix}\right.\)         ⇔  \(\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{9}\\\dfrac{1}{x}=\dfrac{1}{18}\end{matrix}\right.\)      ⇔   \(\left\{{}\begin{matrix}y=9\\x=18\end{matrix}\right.\)

Answer 2

\(Đặt:a=\dfrac{1}{x};b=\dfrac{1}{y}\left(Với:x\ne0;y\ne0\right)\\ \Leftrightarrow\left\{{}\begin{matrix}6a+6b=1\\8a+5b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}30a=30b=5\\48a+30b=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}18a=1\\a+b=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\\dfrac{1}{18}+b=\dfrac{1}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\left(Nhận\right)\\b=\dfrac{1}{9}\left(Nhận\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{a}=18\\y=\dfrac{1}{b}=9\end{matrix}\right.\)