a, \(3\dfrac{1}{2}-\dfrac{2}{3}:\left(1-2x\right)=1\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(1-2x\right)=\dfrac{7}{2}-\dfrac{3}{2}=2\)
\(\Leftrightarrow1-2x=\dfrac{2}{3}:2=\dfrac{1}{3}\Leftrightarrow2x=1-\dfrac{1}{3}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)
b, pt \(\Leftrightarrow\dfrac{-1,5}{3\left|2x+1\right|}=\dfrac{-1}{14}\Rightarrow3\left|2x+1\right|=21\Rightarrow\left|2x+1\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=-7\\2x-1=7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
\(a,3\dfrac{1}{2}-\dfrac{2}{3}:\left(1-2x\right)=1\dfrac{1}{2}\)
\(\Rightarrow\dfrac{2}{3}:\left(1-2x\right)=3\dfrac{1}{2}-1\dfrac{1}{2}\)
\(\Rightarrow\dfrac{2}{3}:\left(1-2x\right)=2\)
\(\Rightarrow1-2x=\dfrac{2}{3}:2\)
\(\Rightarrow1-2x=\dfrac{1}{3}\)
\(\Rightarrow2x=1-\dfrac{1}{3}\)
\(\Rightarrow2x=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{\dfrac{2}{3}}{2}\Rightarrow x=\dfrac{2}{3}.\dfrac{1}{2}\Rightarrow x=\dfrac{1}{3}\)
Vậy \(x=\dfrac{1}{3}\)
\(b,\dfrac{\left(-1\right).5}{3\left|2x+1\right|}=\dfrac{\sqrt{64}}{-112}\)
Từ đẳng thức:
\(\Rightarrow3\left|2x+1\right|.\sqrt{64}=\left[\left(-1\right).5\right].\left(-112\right)\)
\(\Rightarrow3\left|2x+1\right|.8=\left[\left(-1\right).5\right].\left(-112\right)\)
\(\Rightarrow3\left|2x+1\right|.8=\left(-5\right).\left(-112\right)\)
\(\Rightarrow3\left|2x+1\right|.8=560\)
\(\Rightarrow\left|2x+1\right|=\dfrac{560}{\dfrac{8}{7}}\)
\(\Rightarrow\left|2x+1\right|=10\)
Ta có:
\(\left\{{}\begin{matrix}2x+1=10\\2x+1=\left(-10\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10-1}{2}\\x=\dfrac{\left(-10\right)-1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4,5\\x=-5.5\end{matrix}\right.\)
Vậy \(x\in\left\{-5,5;4,5\right\}\)
