Bài 3:
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{6}\\\dfrac{2}{a}+\dfrac{3}{b}=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{a}+\dfrac{3}{b}=\dfrac{1}{2}\\\dfrac{2}{a}+\dfrac{3}{b}=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{10}\\\dfrac{1}{b}=\dfrac{1}{6}-\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=10\\b=15\end{matrix}\right.\)
1.\(\Leftrightarrow\left\{{}\begin{matrix}4x+8y=28\\4x-5y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=28-8y\\28-8y-5y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=28-8y\\13y=26\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=28-16\\y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)