\(a)\left(x+5\right)^2=16\)
\(\Leftrightarrow\left(x+5\right)^2=4^2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
\(b)\left(2x+5\right)^3=8\)
\(\Leftrightarrow\left(2x+5\right)^3=2^3\)
\(\Leftrightarrow2x+5=2\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Lời giải:
a) \((x+5)^2=16=4^2\Rightarrow \left[\begin{matrix} x+5=\sqrt{16}=4\\ x+5=-\sqrt{16}=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=-9\end{matrix}\right.\)
b)
\((2x+5)^3=8\Rightarrow 2x+5=\sqrt[3]{8}=2\)
\(\Rightarrow 2x=-3\Rightarrow x=\frac{-3}{2}\)
\(a)\left(x+5\right)^2=16\)
\(\Leftrightarrow\left(x+5\right)^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x+5=\pm4\)
\(\Leftrightarrow x+5\in\left\{-4;4\right\}\)
\(\Leftrightarrow x\in\left\{-9;-1\right\}\)
a, Ta có \(\left(x+5\right)^2=16\)
Mà \(16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+5=-4\\x+5=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-4-5\\x=4-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-9\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{-9;-1\right\}\)
b, Ta có \(\left(2x+5\right)^3=8\)
\(\Rightarrow\left(2x+5\right)^3=2^3\)
\(\Rightarrow2x+5=2\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=\dfrac{-3}{2}\)
Vậy \(x=\dfrac{-3}{2}\)
