\(a,\left(x-3\right)^2=1\)
=> \(\sqrt{\left(x-3\right)^2}=\sqrt{1}\)
=> \(\left|x-3\right|=1\)
=> \(\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}\)
\(\left(2x+1\right)^3=-8\)
=> \(\sqrt[3]{\left(2x+1\right)^3}=\sqrt[3]{-8}\)
=> \(2x+1=-2\)
=> \(2x=-2-1=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
Vậy \(x\in\left\{-\frac{3}{2}\right\}\)
\(c,\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
=> \(\sqrt{\left(x-\frac{1}{4}\right)^2}=\sqrt{\frac{1}{25}}\)
=> \(\left|x-\frac{1}{4}\right|=\frac{1}{5}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}=\frac{9}{20}\\x=-\frac{1}{5}+\frac{1}{4}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\)
a) ( x-3 )2 = 1
\(\Rightarrow\) (x-3)2 = 12
\(\Rightarrow\) x - 3 = 1
\(\Rightarrow\) x = 4
b) (2x+1)3 = -8
\(\Rightarrow\) (2x+1)3 = (-2)3
\(\Rightarrow\) 2x+1 = -2
\(\Rightarrow\) 2x = -3
\(\Rightarrow\) x = -1,5
c) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
\(\Rightarrow\)\(\left(x-\frac{1}{4}\right)^2=\left(\frac{1}{5}\right)^2\)
\(\Rightarrow\) \(x-\frac{1}{4}=\frac{1}{5}\)
\(\Rightarrow x=\frac{9}{20}\)
\(\left(x-3\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3+1=4\\x=-1+3=2\end{matrix}\right.\)
\(\left(2x+1\right)^3=-8\\ \Leftrightarrow\left(2x+1\right)^3=\left(-2\right)^3\\ \Leftrightarrow2x+1=-2\\ \Leftrightarrow2x=-2-1\\ \Leftrightarrow2x=-3\\ \Leftrightarrow x=-\frac{3}{2}\)
\(\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\\ \Leftrightarrow x-\frac{1}{4}=\pm\frac{1}{5}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}\\x=-\frac{1}{5}+\frac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{20}\\x=\frac{1}{20}\end{matrix}\right.\)
a) \(\left(x-3\right)^2=1\)
=> \(\left\{{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1+3\\x=\left(-1\right)+3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}\).
b) \(\left(2x+1\right)^3=-8\)
=> \(\left(2x+1\right)^3=\left(-2\right)^3\)
=> \(2x+1=-2\)
=> \(2x=\left(-2\right)-1\)
=> \(2x=-3\)
=> \(x=\left(-3\right):2\)
=> \(x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\).
c) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
=> \(x-\frac{1}{4}=\pm\frac{1}{5}\)
=> \(\left\{{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}\\x=\left(-\frac{1}{5}\right)+\frac{1}{4}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{9}{20}\\x=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\).
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