Câu trả lời:
Gọi a=\(\dfrac{1}{x}\), b=\(\dfrac{1}{y}\)
⇔ \(\left\{{}\begin{matrix}6\left(a+b\right)=1\\5\left(a+b\right)+3a=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}6a+6b=1\\5a+5b+3a=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}6a+6b=1\\8a+5b=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}24a+24b=4\\24a+15b=3\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}9b=1\\8a+5b=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{9}\\8a+5.\dfrac{1}{9}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{9}\\a=\dfrac{1}{18}\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{9}\\\dfrac{1}{x}=\dfrac{1}{18}\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}y=9\\x=18\end{matrix}\right.\)