\(2.\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{8}=0\\ 2.\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{8}\\ \left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}=\pm\left(\dfrac{1}{4}\right)^2\)
TH1 : \(x-\dfrac{1}{2}=\dfrac{1}{4}\Rightarrow x=\dfrac{3}{4}\)
TH2 : \(x-\dfrac{1}{2}=-\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}\)
Vậy...
\(2\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{8}=0\)
\(2x^2-\dfrac{1}{4}-\dfrac{1}{8}=0\)
\(2x^2=0+\dfrac{1}{4}+\dfrac{1}{8}\)
\(2x^2=\dfrac{3}{8}\)
\(x^2=\dfrac{3}{8}\div2\)
\(x^2=\dfrac{3}{16}\)
\(x=\sqrt{\dfrac{3}{16}}=\dfrac{\sqrt{3}}{4}\)
\(2\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{8}=0\\ \Leftrightarrow2\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{8}\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{4}\\x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
