a) \(\left(4,5-2x\right):\left(\dfrac{3}{4}\right)=1\dfrac{1}{3}\)
\(\Rightarrow\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{4}{3}=\dfrac{4}{3}\)
\(\Leftrightarrow6-\dfrac{8}{3}x=\dfrac{4}{3}\)
\(\Leftrightarrow18-8x=4\)
\(\Leftrightarrow-8x=4-18\)
\(\Leftrightarrow-8x=-14\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
b) \(\dfrac{1}{2}-\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{2}{3}x=\dfrac{2}{3}\)
\(\Leftrightarrow5-4x=4\)
\(\Leftrightarrow-4x=4-5\)
\(\Leftrightarrow-4x=-1\)
\(\Rightarrow x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
a) \(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(4,5-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(4,5-2x=\dfrac{4}{3}.\dfrac{3}{4}\)
\(4,5-2x=1\)
\(2x=4,5-1\)
\(2x=3,5=\dfrac{7}{2}\)
\(x=\dfrac{7}{2}:2\)
\(x=\dfrac{7}{4}\)
b) \(\dfrac{1}{2}-\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{-1}{6}\)
\(\dfrac{2}{3}x=\dfrac{-1}{6}+\dfrac{1}{3}\)
\(\dfrac{2}{3}x=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}:\dfrac{2}{3}\)
\(x=\dfrac{1}{4}\)
a) (4,5-2x):\(\dfrac{3}{4}=1\dfrac{1}{3}\)
(4,5-2x)\(\dfrac{3}{4}=\dfrac{4}{3}\)
4,5-2x=\(\dfrac{4}{3}.\dfrac{3}{4}\)
4,5-2x=1
2x=4,5-1
2x=3,5
x=3,5:2
x=\(\dfrac{7}{4}\)
b)\(1\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{2}{3}:1\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3}x=\dfrac{2}{3}+\dfrac{1}{3}\)
\(\dfrac{2}{3}x=1\)
x=\(1:\dfrac{2}{3}=\dfrac{3}{2}\)
a, \(\left(4,5-2x\right)\div\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\dfrac{9}{2}-2x=\dfrac{4}{3}.\dfrac{3}{4}\)
\(\dfrac{9}{2}-2x=1\Rightarrow2x=\dfrac{9}{2}-1\)
\(\Rightarrow2x=\dfrac{7}{2}\Rightarrow x=\dfrac{7}{2}\div2=\dfrac{7}{4}\)
Vậy x = \(\dfrac{7}{4}\)
b, \(\dfrac{1}{2}-\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{-1}{6}\)
\(\dfrac{2}{3}x=\dfrac{-1}{6}+\dfrac{1}{3}\Rightarrow\dfrac{2}{3}x=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}\div\dfrac{2}{3}=\dfrac{1}{6}.\dfrac{3}{2}=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
