Bài 10:
a) Ta có: \(\dfrac{-3}{2}-2x+\dfrac{3}{4}=-2\)
\(\Leftrightarrow\dfrac{-3}{4}-2x=-2\)
\(\Leftrightarrow2x=\dfrac{-3}{4}+2=\dfrac{5}{4}\)
hay \(x=\dfrac{5}{8}\)
b) Ta có: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(\dfrac{3}{-2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-12}{145}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{15}{29}\)
hay \(x=-\dfrac{45}{58}\)
c) Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7}{10}x\)
\(\Leftrightarrow\dfrac{-1}{10}x+\dfrac{7}{10}x=\dfrac{-7}{5}-\dfrac{13}{5}=-2\)
\(\Leftrightarrow\dfrac{3}{5}x=-2\)
hay \(x=-2:\dfrac{3}{5}=\dfrac{-10}{3}\)
d) Ta có: \(\dfrac{2x-3}{3}-\dfrac{3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow2\left(2x-3\right)-3=5-3x-2\)
\(\Leftrightarrow4x-6-3+3x-3=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\dfrac{12}{7}\)
e) Ta có: \(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
\(\Leftrightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{1}{4}+\dfrac{4}{5}+2\)
\(\Leftrightarrow\dfrac{2}{3x}+\dfrac{21}{3x}=\dfrac{61}{20}\)
\(\Leftrightarrow3x=\dfrac{460}{61}\)
hay \(x=\dfrac{460}{183}\)
