\(a,\dfrac{-3}{4}x+1=\dfrac{5}{6}\\ \Rightarrow\dfrac{-3}{4}x=\dfrac{-1}{6}\\ \Rightarrow x=\dfrac{2}{9}\\ b,\left(2x-3\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\\ c,\dfrac{1}{2}-\left|x+1\right|=0,25\\ \Rightarrow\left|x+1\right|=0,25\\ \Rightarrow\left[{}\begin{matrix}x+1=0,25\\x+1=-0,25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-0,75\\x=-1,25\end{matrix}\right.\)
a: =>-3/4x=-1/6
hay x=2/9
b: =>2x-3=0 hoặc x-5=0
hay x=3/2 hoặc x=5
c: =>|x+1|=1/4
\(\Leftrightarrow x+1\in\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
hay \(x\in\left\{-\dfrac{3}{4};-\dfrac{5}{4}\right\}\)
b, \(=>2x-3=0\) hay \(x-5=0\)
\(2x=0+3\) \(x=0+5\)
\(2x=3\) \(x=5\)
\(x=\dfrac{3}{2}\)
Vậy x ϵ { \(\dfrac{3}{2}\);5}
a, \(\dfrac{-3}{4}x=\dfrac{5}{6}-1\)
\(\dfrac{-3}{4}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{-3}{4}\)
\(x=\dfrac{2}{9}\)
