Ta có:
(2-\(\dfrac{5}{6}x\)):(-2)+\(\dfrac{1}{3}x\)=-0,91
=>2:(-2)+\(\dfrac{5}{6}x:2\)+\(\dfrac{1}{3}x\)=-0,91
=>(-1)+\(\dfrac{5}{12}x\)+\(\dfrac{1}{3}x\)=-0,91
=>\(\left(\dfrac{5}{12}+\dfrac{1}{3}\right)x=\left(-0,91\right)+1\)
=>\(\dfrac{3}{4}x=0,09\)
=>\(x=0,09:\dfrac{3}{4}\)
=>\(x=\dfrac{9}{100}\cdot\dfrac{4}{3}\)
\(=>x=\dfrac{3}{25}\)
\(\left(2-\dfrac{5}{6}x\right):\left(-2\right)+\dfrac{1}{3}x=-0,91\)
\(\Leftrightarrow\dfrac{2}{-2}-\dfrac{\dfrac{5}{6}x}{-2}+\dfrac{1}{3}x=-0,91\)
\(\Leftrightarrow-1+\dfrac{5}{12}x+\dfrac{1}{3}x=-0,91\)
\(\Leftrightarrow-1+x\left(\dfrac{5}{12}+\dfrac{1}{3}\right)=-0,91\)
\(\Leftrightarrow-1+\dfrac{3}{4}x=-0,91\)
\(\Leftrightarrow\dfrac{3}{4}x=-0,91-\left(-1\right)\)
\(\Leftrightarrow\dfrac{3}{4}x=-0,91+1\)
\(\Leftrightarrow\dfrac{3}{4}x=-\left(0,91-1\right)\)
\(\Leftrightarrow\dfrac{3}{4}x=-\left(-0,09\right)\)
\(\Leftrightarrow\dfrac{3}{4}x=0,09\)
\(\Leftrightarrow x=0,09:\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{25}\)
Vậy \(x=\dfrac{3}{25}\)
