\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{-1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\) vậy \(x=\dfrac{-1}{4};x=\dfrac{-3}{4}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2=\left(-\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{\dfrac{1}{4};\dfrac{-1}{4}\right\}\)
TH1:\(x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
TH2:\(x+\dfrac{1}{2}=-\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}-\dfrac{1}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
Vậy \(x\in\left\{\dfrac{-3}{4};\dfrac{-1}{4}\right\}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\pm\dfrac{1}{4}^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\left(x+\dfrac{1}{2}\right)=\dfrac{1}{4}\)
\(x=-\dfrac{1}{4}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\left(x+\dfrac{1}{2}\right)^2=\pm\dfrac{1}{4}^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\pm\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
