a) \(x+\dfrac{7}{-2}=\dfrac{3}{4}\)
\(x-\dfrac{7}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{7}{2}\)
\(x=\dfrac{17}{4}\)
b) \(9x-\dfrac{14}{9}=-\dfrac{2}{9}\)
\(9x=\dfrac{14}{9}-\dfrac{2}{9}\)
\(9x=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}\div9\)
\(x=\dfrac{4}{27}\)
c) \(-\dfrac{1}{2}\div\left(x-\dfrac{1}{3}\right)=-\dfrac{3}{4}\)
\(x-\dfrac{1}{3}=-\dfrac{1}{2}\div\dfrac{-3}{4}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{1}{3}\)
\(x=1\)
d) \(\left(x-3,5\right)\div3\dfrac{1}{2}-2,5=-1\dfrac{3}{4}\)
\(\left(x-\dfrac{7}{2}\right)\div\dfrac{7}{2}-\dfrac{5}{2}=-\dfrac{7}{4}\)
\(\left(x-\dfrac{7}{2}\right)\div\dfrac{7}{2}=\dfrac{5}{2}-\dfrac{7}{4}\)
\(\left(x-\dfrac{7}{2}\right)\div\dfrac{7}{2}=\dfrac{3}{4}\)
\(x-\dfrac{7}{2}=\dfrac{3}{4}.\dfrac{7}{2}\)
\(x-\dfrac{7}{2}=\dfrac{21}{8}\)
\(x=\dfrac{21}{8}+\dfrac{7}{2}\)
\(x=\dfrac{49}{8}\)
a) \(x+\dfrac{7}{-2}=\dfrac{3}{4}\\x=\dfrac{3}{4}-\dfrac{7}{-2}\\ x=\dfrac{3}{4}+\dfrac{7}{2}\\ x=\dfrac{3}{4} +\dfrac{14}{4}=\dfrac{17}{4}\)
a) Ta có: \(x-\dfrac{7}{2}=\dfrac{3}{4}\)
nên \(x=\dfrac{17}{4}\)
b) Ta có: \(9x-\dfrac{14}{9}=\dfrac{-2}{9}\)
\(\Leftrightarrow9x=\dfrac{-2}{9}+\dfrac{14}{9}=\dfrac{4}{3}\)
hay \(x=\dfrac{4}{27}\)
c) Ta có: \(\dfrac{-1}{2}:\left(x-\dfrac{1}{3}\right)=\dfrac{-3}{4}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{-1}{2}:\dfrac{-3}{4}=\dfrac{-1}{2}\cdot\dfrac{4}{-3}=\dfrac{2}{3}\)
hay x=1
d) Ta có: \(\left(x-3.5\right):3\dfrac{1}{2}-2.5=-1\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}=\dfrac{-7}{4}+\dfrac{5}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow x-\dfrac{7}{2}=\dfrac{3}{4}\cdot\dfrac{7}{2}=\dfrac{21}{8}\)
hay \(x=\dfrac{21}{8}+\dfrac{7}{2}=\dfrac{21}{8}+\dfrac{28}{8}=\dfrac{49}{8}\)
