a,\(\left(2x-3\right)^2=16\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
b,\(\left(x-\frac{2}{3}\right)^3=\frac{1}{27}\Leftrightarrow x-\frac{2}{3}=\frac{1}{3}\Leftrightarrow x=1\)
Bài 1:
a) \(\left(2x-3\right)^2=16\)
\(\Rightarrow2x-3=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4+3=7\\2x=\left(-4\right)+3=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7:2\\x=\left(-1\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{1}{2}\right\}.\)
b) \(\left(x-\frac{2}{3}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}+\frac{2}{3}\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
Chúc bạn học tốt!
\( a) (2x-3)^2=16\\ (2x-3)^2 = 4^2 hay (2x-3)^2 = (-4)^2 \\ 2x-3 = 4 hay 2x-3 = -4\\ 2x = 4+3 hay 2x = -4 +3\\ 2x = 7 hay 2x=-1\\ x=\frac{7}{2} hay x = \dfrac {-1}{2}\\ Vậy x=\frac{7}{2} hay x = \dfrac {-1}{2} \)
\(b) \bigg(x-\frac{2}{3}\bigg )^3 = \dfrac{1}{27}\\ \bigg(x-\frac{2}{3}\bigg )^3 =\bigg ( \dfrac{1}{3}\bigg)^3 \\ x-\frac{2}{3}= \dfrac{1}{3}\\ x = \dfrac {1}{3}+\dfrac{2}{3}\\ x=1\\ Vậy x=1 \)
