\(\left(2x+1\right)^3=-8\\ \left(2x+1\right)^3=\left(-2\right)^3\\ \Rightarrow2x+1=-2\\ \Rightarrow2x=-3\\ \Rightarrow x=\frac{-3}{2}\)
Vậy \(x=\frac{-3}{2}\)
\(\left(3x+2\right)^2=16\\ \left(3x+2\right)^2=4^2=\left(-4\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{2}{3};-2\right\}\)
\(\left(2x+1\right)^3=-8\)
⇔ \(\left(2x+1\right)^3=\left(-2\right)^3\)
⇔ \(2x+1=-2\)
⇔ \(2x=\left(-2\right)-1\)
⇔ \(2x=-3\)
⇔ \(x=\left(-3\right):2\)
=> \(x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\).
\(\left(3x+2\right)^2=16\)
⇔ \(3x+2=\pm4\)
⇔ \(\left\{{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}3x=4-2=2\\3x=\left(-4\right)-2=-6\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=2:3\\x=\left(-6\right):3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{2}{3};-2\right\}\).
Chúc bạn học tốt!
\({\left( {2x + 1} \right)^3} = - 8\\ \Leftrightarrow {\left( {2x + 1} \right)^3} = {\left( { - 2} \right)^3}\\ \Leftrightarrow 2x + 1 = - 2\\ \Leftrightarrow 2x = - 2 - 1\\ \Leftrightarrow 2x = - 3\\ \Leftrightarrow x = - \dfrac{3}{2} \)
\( {\left( {3x + 2} \right)^2} = 16\\ \Leftrightarrow 3x + 2 = \pm 4\\ \Leftrightarrow \left[ \begin{array}{l} 3x + 2 = 4\\ 3x + 2 = - 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 3x = 4 - 2\\ 3x = - 4 - 2 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 3x = 2\\ 3x = - 6 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{2}{3}\\ x = \dfrac{{ - 6}}{3} = - 2 \end{array} \right. \)
