a) \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
\(\Leftrightarrow x=0+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
a, \(\left(x-\dfrac{3}{4}\right)^2=0\Rightarrow x=\dfrac{3}{4}\)
Vậy...
b, \(\left(x-3\right)^2=1\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy x = 4 hoặc x = 2
c, \(\left(2x+1\right)^3=-8\)
\(\Rightarrow2x+1=-3\)
\(\Rightarrow x=-2\)
Vậy x = -2
d, \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{2}\\x-\dfrac{1}{4}=\dfrac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy...
c) \(\left(2x+1\right)^3=-8\)
\(\Leftrightarrow\left(2x+1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x+1=-2\)
\(\Leftrightarrow2x=\left(-2\right)-1\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
d) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x-\dfrac{1}{4}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{2}\\x-\dfrac{1}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{4}\\x=\left(-\dfrac{1}{2}\right)+\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left(x-3\right)^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
