a) x tử hay mẫu cần rõ ra nhé
b) \(\dfrac{2}{3}+\dfrac{1}{3}:x=-1\Leftrightarrow\dfrac{1}{3x}=-1-\dfrac{2}{3}=-\dfrac{5}{3}\Leftrightarrow\dfrac{1}{x}=-5;x=\dfrac{-1}{5}\)
\(\left|2x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5;2x=6;x=3\\2x-1=-5;2x=-4;x-2\end{matrix}\right.\)
\(a,\dfrac{3}{4}x+1=-2\)
\(\dfrac{3}{4}x=-2-1\)
\(\dfrac{3}{4}x=-3\)
\(x=\dfrac{-3}{\dfrac{3}{4}}=-4\)
Vậy \(x=-4\)
\(b,\dfrac{2}{3}+\dfrac{1}{3}:x=-1\)
\(\dfrac{1}{3}:x=-1-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{5}{3}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{5}{3}\right)=-\dfrac{1}{5}\)
Vậy \(x=-\dfrac{1}{5}\)
\(c,\left|2x-1\right|=5\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5+1}{2}\\x=\dfrac{-5+1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;3\right\}\)
