Question

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

Answer 1

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

\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)

\(\Leftrightarrow4x-15=-3x+3\)

\(\Leftrightarrow4x+3x=3+15\)

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

Answer 2

\(\Leftrightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}\)

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

\(\Leftrightarrow4x-6-5+3x=7\)

\(\Leftrightarrow7x=7+5+6=18\)

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

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