a) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)
b) \(\left(3x+1\right)^3=-27\)
\(\Rightarrow\left(3x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+1=-3\)
\(\Rightarrow3x=\left(-3\right)-1\)
\(\Rightarrow3x=-4\)
\(\Rightarrow x=\left(-4\right):3\)
\(\Rightarrow x=-\frac{4}{3}\)
Vậy \(x=-\frac{4}{3}.\)
Mấy câu sau làm tương tự nhé.
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c)\(\left(3x-2\right)^2=36\\ \Leftrightarrow\left(3x-2\right)^2=\left(\pm6\right)^2\\ \Rightarrow\left\{{}\begin{matrix}3x-2=6\\3x-2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{4}{3}\end{matrix}\right.\)
d)\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\\ \Leftrightarrow\left(\frac{2}{5}-3x\right)^2=\left(\pm\frac{3}{5}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\frac{2}{5}-3x=\frac{3}{5}\\\frac{2}{5}-3x=-\frac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{15}\\x=\frac{1}{3}\end{matrix}\right.\)