Question

Answer 1

$\dfrac{2x-3}{3}+\dfrac{-3}{x}=\dfrac{5-3x}{6}-\dfrac{1}{3}\\\Leftrightarrow 2(2x-3)-9=5-3x-2\\\Leftrightarrow 4x-6-9=3-3x\\\Leftrightarrow 4x-15=3-3x\\\Leftrightarrow 7x=18\\\Leftrightarrow x=\dfrac{18}{7}$

Vậy `S={18/7}`

Answer 2

\(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\) 

\(\Rightarrow\dfrac{2.\left(2x-3\right)+\left(-3\right).3}{6}=\dfrac{5-3x-2}{6}\) 

\(\Leftrightarrow4x-6-9=5-2-3x\) 

\(\Leftrightarrow4x+3x=5-2+6+9\) 

\(\Leftrightarrow7x=18\) 

\(\Leftrightarrow x=18:7\) 

\(\Leftrightarrow x=\dfrac{18}{7}\)