a) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\dfrac{-2}{3}x\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}x\)
\(x-\dfrac{2}{3}x=\dfrac{2}{5}-\dfrac{5}{6}\)
\(\dfrac{1}{3}x=-\dfrac{13}{30}\)
\(x=-1,3\)
b) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}-\dfrac{1}{5}x\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}-\dfrac{1}{5}x\)
\(\dfrac{7}{5}x+\dfrac{1}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{8}{5}x=-\dfrac{43}{35}\)
\(x=-\dfrac{43}{56}\)
c)\(\dfrac{1}{3}-\dfrac{2}{3}\left(x-\dfrac{1}{2}\right)=1\)
\(\dfrac{1}{3}-\dfrac{2}{3}x+\dfrac{1}{3}=1\)
\(-\dfrac{2}{3}x=1-\dfrac{1}{3}-\dfrac{1}{3}\)
\(-\dfrac{2}{3}x=\dfrac{1}{3}\)
\(x=-\dfrac{1}{2}\)
a)x+\(\dfrac{5}{6}\)=\(\dfrac{2}{3}-\dfrac{-2}{3}x\)
=>x+\(\dfrac{2}{3}x\)=\(\dfrac{2}{3}\)-\(\dfrac{5}{6}\)
=>\(\dfrac{5}{3}x\)=\(\dfrac{-1}{6}\)
=>\(x\)=\(\dfrac{-1}{10}\)
Vậy \(x=\dfrac{-1}{10}\)
b)1\(\dfrac{2}{5}x\)+\(\dfrac{3}{7}\)=\(\dfrac{-4}{5}-\dfrac{1}{5}x\)
=>\(\dfrac{7}{5}x\)+\(\dfrac{1}{5}x\)=\(\dfrac{-4}{5}-\dfrac{3}{7}\)
=>\(\dfrac{8}{5}x\)=\(\dfrac{-43}{35}\)
=>\(x\)=\(\dfrac{-43}{56}\)
KL:...
c)\(\dfrac{1}{3}\)−\(\dfrac{2}{3}\)(x−\(\dfrac{1}{2}\))=1
=>\(\dfrac{-2}{3}\)\(\left(x-\dfrac{1}{2}\right)\)=1-\(\dfrac{1}{3}\)
=>\(\dfrac{-2}{3}x\)+\(\dfrac{1}{3}\)=\(\dfrac{2}{3}\)
=>\(\dfrac{-2}{3}x\)=\(\dfrac{1}{3}\)
=>\(x\)=\(\dfrac{-1}{2}\)
KL: