a, \(\frac{-2}{3}x=\frac{4}{15}\)
x = \(\frac{4}{15}:\frac{-2}{3}\)
x = \(\frac{-2}{5}\)
b, x + \(\frac{1}{3}\) = \(\frac{2}{5}-\left(\frac{-1}{3}\right)\)
x = \(\frac{11}{15}-\frac{1}{3}\)
x = \(\frac{2}{5}\)
c, \(\frac{-2}{3}\)x + \(\frac{-3}{7}\) + \(\frac{1}{2}\)x = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x + \(\frac{-3}{7}\) = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x = \(\frac{-5}{6}+\frac{3}{7}\)
\(\frac{-1}{6}\)x = \(\frac{-17}{42}\)
x = \(\frac{-17}{42}:\frac{-1}{6}\)
x = \(\frac{17}{7}\)
d, 2x(x - \(\frac{1}{7}\)) = 0
\(\Rightarrow\) 2x = 0 hoặc x - \(\frac{1}{7}\) = 0
TH1: 2x = 0
x = 0
TH2: x - \(\frac{1}{7}\) = 0
x = \(\frac{1}{7}\)
Vậy x = 0; x = \(\frac{1}{7}\)
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