a) \(5x-1=2x-4\)
\(3x+3=0\)
\(3x=-3\)
\(x=-1\)
Vậy \(x=-1\)
b) \(\left(2x-1\right)^2-\dfrac{25}{4}=0\)
\(\left(2x-1\right)^2=\dfrac{25}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{5}{4}\right)^2=\left(-\dfrac{5}{4}\right)^2\)
TH1:
\(2x-1=\dfrac{5}{4}\)
\(2x=\dfrac{9}{4}\)
\(x=\dfrac{9}{8}\)
TH2:
\(2x-1=-\dfrac{5}{4}\)
\(2x=-\dfrac{1}{4}\)
\(x=-\dfrac{1}{8}\)
Vậy \(x\in\left\{-\dfrac{1}{8};\dfrac{9}{8}\right\}\)
a) Ta có: 5x-1=2x-4
nên 5x-2x=-4+1
\(\Leftrightarrow3x=-3\)
hay x=-1
b) Ta có: \(\left(2x-1\right)^2-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(2x-1-\dfrac{5}{2}\right)\left(2x-1+\dfrac{5}{2}\right)=0\)
\(\Leftrightarrow\left(2x-\dfrac{7}{2}\right)\left(2x+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{2}\\2x=\dfrac{-3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
