a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).
a , \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b , \(\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)
d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)
<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=0+\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=\left(-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) \(\left(TM\right)\)
Vậy \(x\in\left\{3;1\right\}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-2+1\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
Vậy..........
b/ \(\left(x-2\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy...............
c/ \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-4\right)^3\)
\(\Rightarrow2x-1=-4\)
\(\Rightarrow x=\dfrac{-4+1}{2}=-\dfrac{3}{2}\)
Vậy...........
d/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy........
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