Question

Answer 1

PT=>2(2x-3)-9=5-3x-2

=>4x-6-9=-3x+3

=>4x-15=-3x+3

=>7x=18

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

Answer 2

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

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

\(\Rightarrow\dfrac{2\left(2x-3\right)}{6}-\dfrac{5-3x}{6}=\dfrac{7}{6}\)

\(\Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\)

\(\Rightarrow\dfrac{7x-11}{6}=\dfrac{7}{6}\)

\(\Rightarrow7x-11=7\)

\(\Rightarrow7x=18\)

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

\(Toru\)

Answer 3

\(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\\ \Rightarrow\dfrac{2\left(2x-3\right)-3.3-\left(5-3x\right)+2}{6}=0\\ \Rightarrow4x-6-9-5+3x+2=0\)

\(\Rightarrow7x=18\\ \Rightarrow x=\dfrac{18}{7}\)

Vậy ...