2.
\(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\\ \text{⇔}-\dfrac{1}{15}+3x=\dfrac{3}{4}\\ \text{⇔}3x=\dfrac{3}{4}+\dfrac{1}{15}=\dfrac{49}{60}\\ \text{⇔}x=\dfrac{49}{180}\)
4.
\(\dfrac{3}{2}-4\left(\dfrac{1}{4}-x\right)=\dfrac{2}{3}-7x\\ \text{⇔}\dfrac{3}{2}-1+4x=\dfrac{2}{3}\\ \text{⇔}4x+\dfrac{1}{2}=\dfrac{2}{3}\\ \text{⇔}4x=\dfrac{1}{6}\\ \text{⇔}x=\dfrac{1}{24}\)
6.
\(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\\ \text{⇔}2-4x-5x+\dfrac{15}{10}=\dfrac{7}{4}\\ \text{⇔}\dfrac{7}{2}-9x=\dfrac{7}{4}\\ \text{⇔}9x=\dfrac{7}{4}\\ \text{⇔}x=\dfrac{7}{36}\)
8.
\(3-2x-\dfrac{1}{3}=7x-\dfrac{1}{4}\\ \text{⇔}9x-\dfrac{35}{12}=0\\ \text{⇔}x=\dfrac{35}{108}\)
10.
\(-\dfrac{15}{2}+\dfrac{1}{4}+4x-2=1\\ \text{⇔}4x=\dfrac{41}{4}\\ \text{⇔}x=\dfrac{41}{16}\)
12.
\(\dfrac{2}{5}-\dfrac{1}{3}x+\dfrac{1}{6}=\dfrac{1}{4}\\ \text{⇔}\dfrac{17}{30}-\dfrac{1}{3}x=\dfrac{1}{3}\\ \text{⇔}\dfrac{1}{3}x=\dfrac{7}{30}\\ \text{⇔}x=\dfrac{7}{10}\)
14.
\(-1+\dfrac{2}{3}x=\dfrac{1}{8}-\dfrac{1}{6}\\ \text{⇔}\dfrac{2}{3}x=\dfrac{23}{24}\\ \text{⇔}x=\dfrac{23}{16}\)
2) Ta có: \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{3}{4}+\dfrac{2}{5}-\dfrac{1}{3}=\dfrac{45}{60}+\dfrac{24}{60}-\dfrac{20}{60}=\dfrac{49}{60}\)
hay \(x=\dfrac{49}{180}\)
4) Ta có: \(\dfrac{3}{2}-4\left(\dfrac{1}{4}-x\right)=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{3}{2}-1+4x-\dfrac{2}{3}+7x=0\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
hay \(x=\dfrac{1}{66}\)
6) Ta có: \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x+\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=-\dfrac{7}{4}\)
hay \(x=\dfrac{7}{36}\)
8) Ta có: \(2\left(\dfrac{3}{2}-x\right)-\dfrac{1}{3}=7x-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x-\dfrac{1}{3}-7x+\dfrac{1}{4}=0\)
\(\Leftrightarrow-9x=-\dfrac{35}{12}\)
hay \(x=\dfrac{35}{108}\)
10) Ta có: \(-\dfrac{3}{2}\left(5-\dfrac{1}{6}\right)+4\left(x-\dfrac{1}{2}\right)=1\)
\(\Leftrightarrow\dfrac{-15}{2}+\dfrac{1}{4}+4x-2=1\)
\(\Leftrightarrow4x=\dfrac{41}{4}\)
hay \(x=\dfrac{41}{16}\)
12) Ta có: \(\dfrac{2}{5}-\dfrac{1}{3}\left(x-\dfrac{1}{2}\right)=\dfrac{1}{2}-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{1}{3}x+\dfrac{1}{6}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{3}x=-\dfrac{19}{60}\)
hay \(x=\dfrac{19}{20}\)