\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{2}\\x-\dfrac{1}{3}=\dfrac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
Vậy ...
\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
<=>\(\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\)
\(x-\dfrac{1}{3}=\dfrac{1}{2}\Rightarrow x=\dfrac{5}{6}\)
\(x-\dfrac{1}{3}=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{5}{6}\) hoặc \(x=\dfrac{-1}{6}\)
Học tốt nhé
\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)
\(\left(x-\dfrac{1}{3}\right)^2=0+\dfrac{1}{4}\)
\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{5}{6}\)
\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{2}\\x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
Vậy...
tik mik nha !!!
