\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)
\(b,\frac{1}{4}:x=-3+\frac{3}{4}\)<=>\(\frac{1}{4}:x=\frac{-9}{4}\)<=>\(x=-9\)
\(c,\orbr{\begin{cases}2x-\frac{1}{3}=0\\0.5x+0.25=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-3}{4}\end{cases}}\)
a) \(\frac{-2}{3}x+4=12\)
\(\Rightarrow\frac{-2}{3}x=12-4\)
\(\Rightarrow\frac{-2}{3}x=8\)
\(\Rightarrow x=8:\frac{-2}{3}\)
\(\Rightarrow x=-12\)
Vậy x = -12
b) \(\frac{-3}{4}+\frac{1}{4}:x=-3\)
\(\Rightarrow\frac{1}{4}:x=-3-\left(\frac{3}{4}\right)\)
\(\Rightarrow\frac{1}{4}:x=\frac{-9}{4}\)
\(\Rightarrow x=\frac{1}{4}:\frac{-9}{4}\)
\(\Rightarrow x=\frac{-1}{9}\)
Vậy \(x=\frac{-1}{9}\)
c) \(\left(2x-\frac{1}{3}\right)\left(0,5x+0,25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\0,5x+0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0,5x=-0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-0,5\end{cases}}\)
Vậy \(x=\frac{1}{6}\)hoặc \(x=-0,5\)
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