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a, \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{3}{2}x=\frac{-5}{4}\)
\(\Leftrightarrow x=\frac{-5}{6}\)
b, \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
\(\Leftrightarrow\frac{3}{5}.\left(3x-\frac{37}{10}\right)=\frac{-57}{10}\)
\(\Leftrightarrow3x-\frac{37}{10}=\frac{-19}{2}\)
\(\Leftrightarrow3x=\frac{-29}{5}\)
\(\Leftrightarrow x=\frac{-29}{15}\)
a) \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=-\frac{3}{4}\)
\(x\left(\frac{2}{3}+\frac{5}{6}\right)+\frac{1}{2}=-\frac{3}{4}\)
\(x\cdot\frac{3}{2}=\frac{-5}{4}\)
\(x=-\frac{5}{6}\)
\(\frac{2}{5}+\frac{3}{5}\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\left(3x-\frac{37}{10}\right)=-\frac{57}{10}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=\frac{-29}{5}\)
\(x=\frac{-29}{15}\)
a, \(\frac{3}{7}=-\frac{x}{21}\Leftrightarrow63=-7x\Leftrightarrow x=-9\)
b, \(\frac{x-1}{6}=\frac{2}{3}\Leftrightarrow\frac{x-1}{6}=\frac{4}{6}\Leftrightarrow x-1=4\Leftrightarrow x=5\)
c, \(\frac{-15}{2x+3}=\frac{-5}{8}\Leftrightarrow-120=-10x-15\Leftrightarrow-10x=-105\Leftrightarrow x=10,5\)( chưa xem lại ko chắc , thử lại giúp mk nha )
a) \(\frac{3}{7}=\frac{-x}{21}\)
\(\Rightarrow-7x=21.3\)
\(-7x=63\)
\(x=63:\left(-7\right)\)
\(x=-9\)
Vậy x=-9
b) \(\frac{x-1}{6}=\frac{2}{3}\)
\(\Rightarrow\left(x-1\right).3=6.2\)
\(3x-3=12\)
\(3x=12+3\)
\(3x=15\)
\(x=15:3\)
\(x=5\)
Vậy x=5
\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)
\(\frac{x}{15}=\frac{-1}{15}\)
=> \(x=-1\)
\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)
\(\frac{x}{182}=\frac{-15}{91}\)
=> \(91x=182\cdot\left(-15\right)\)
=> \(91x=-2730\)
=> \(x=-30\)
g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)
\(\Leftrightarrow x=-1\)
h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)
\(\Leftrightarrow x=-30\)
a, \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\Leftrightarrow\frac{x}{3}+\frac{7}{12}=0\Leftrightarrow\frac{4x}{12}+\frac{7}{12}=0\)
Khử mẫu ta đc : \(4x+7=0\Leftrightarrow4x=-7\Leftrightarrow x=-\frac{7}{4}\)
b, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow\frac{x+3}{15}=\frac{5}{15}\)
Khử mẫu ta đc : \(x+3=5\Leftrightarrow x=2\)