Bài 2:
a) Ta có: \(x+3.4=4.5\cdot1.3\)
\(\Leftrightarrow x+\dfrac{17}{5}=\dfrac{9}{2}\cdot\dfrac{13}{10}=\dfrac{117}{20}\)
\(\Leftrightarrow x=\dfrac{117}{20}-\dfrac{17}{5}=\dfrac{117}{20}-\dfrac{68}{20}\)
hay \(x=\dfrac{49}{20}\)
Vậy: \(x=\dfrac{49}{20}\)