\(a,?\\ b,\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\\ =>\dfrac{3}{7}x+1=\dfrac{-1}{28}\cdot\left(-4\right)=\dfrac{1}{7}\\ =>\dfrac{3}{7}x=\dfrac{1}{7}-1=\dfrac{-6}{7}\\ =>x=\dfrac{-6}{7}:\dfrac{3}{7}=-2\\ c,3\left(3x-\dfrac{1}{2}+\dfrac{1}{9}\right)=0\\ =>3x-\dfrac{1}{2}+\dfrac{1}{9}=0\\ =>3x+\dfrac{-7}{18}=0\\ =>3x=\dfrac{7}{18}\\ =>x=\dfrac{7}{18}:3\\ =>x=\dfrac{7}{54}\\ d,\dfrac{1}{2}x+\dfrac{2}{3}\left(x-1\right)=\dfrac{1}{3}\\ =>\dfrac{1}{2}x+\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{1}{3}\\ =>\dfrac{1}{2}x+\dfrac{2}{3}x=\dfrac{1}{3}+\dfrac{2}{3}\\ =>\dfrac{7}{6}x=1\\ =>x=1:\dfrac{7}{6}=\dfrac{6}{7}\\ e,\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ =>\dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ =>\dfrac{1}{3}x+\dfrac{2}{3}x=\dfrac{2}{3}\\ =>x=\dfrac{2}{3}\)
\(a,\dfrac{17}{2}-\left(2x.\dfrac{-3}{4}\right)=\dfrac{-7}{4}\)
\(2x.\dfrac{-3}{4}=\dfrac{17}{2}-\dfrac{-7}{4}\)
\(2x.\dfrac{-3}{4}=\dfrac{34}{4}+\dfrac{7}{4}\)
\(2x.\dfrac{-3}{4}=\dfrac{41}{4}\)
\(2x=\dfrac{41}{4}:\dfrac{-3}{4}\)
\(2x=\dfrac{41}{4}.\dfrac{-4}{3}\)
\(2x=\dfrac{41.\left(-1\right)}{1.3}\)
\(2x=\dfrac{-41}{3}\)
\(x=\dfrac{-41}{3}:2\)
\(x=\dfrac{-41}{3}.\dfrac{1}{2}\)
\(x=\dfrac{-41}{6}\)
Vậy \(x=\dfrac{-41}{6}\)
\(b,\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\dfrac{3x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{\left(-1\right).\left(-1\right)}{7.1}\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(3x.7=\left(-6\right).7\)
\(3x.7=-42\)
\(3x=\left(-42\right):7\)
\(3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
Vậy \(x=-2\)
\(c,3\left(3x-\dfrac{1}{2}+\dfrac{1}{9}\right)=0\)
\(3x-\dfrac{1}{2}+\dfrac{1}{9}=0:3\)
\(3x-\dfrac{1}{2}+\dfrac{1}{9}=0\)
\(3x-\dfrac{1}{2}=0-\dfrac{1}{9}\)
\(3x-\dfrac{1}{2}=\dfrac{-1}{9}\)
\(3x=\dfrac{-1}{9}+\dfrac{1}{2}\)
\(3x=\dfrac{-2}{18}+\dfrac{9}{18}\)
\(3x=\dfrac{7}{18}\)
\(x=\dfrac{7}{18}:3\)
\(x=\dfrac{7}{18}.\dfrac{1}{3}\)
\(x=\dfrac{7}{54}\)
Vậy \(x=\dfrac{7}{54}\)
\(d,\dfrac{1}{2}x+\dfrac{2}{3}\left(x-1\right)=\dfrac{1}{3}\)
\(\dfrac{1}{2}x+\dfrac{2}{3}x-\dfrac{2}{3}.1=\dfrac{1}{3}\)
\(x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x\left(\dfrac{3}{6}+\dfrac{4}{6}\right)=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x\dfrac{7}{6}=\dfrac{3}{3}\)
\(x\dfrac{7}{6}=1\)
\(x=1:\dfrac{7}{6}\)
\(x=\dfrac{6}{7}\)
Vậy \(x=\dfrac{6}{7}\)
\(e,\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\)
\(\dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}.1=0\)
\(x\left(\dfrac{1}{3}+\dfrac{2}{3}\right)-\dfrac{2}{3}=0\)
\(x\dfrac{3}{3}=0+\dfrac{2}{3}\)
\(x1=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}:1\)
\(x=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)
