`5)`\(x^3-3x^2+3x-1=-64\)
\(\Leftrightarrow\left(x-1\right)^3=-64\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-4\right)^3\)
\(\Leftrightarrow x-1=-4\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
`6)`\(x^2+6x+9=\left(x-4\right)^2\)
\(\Leftrightarrow\left(x+3\right)^2=\left(x-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x-4\\x+3=4-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3=-4\left(vô.lý\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
`7)`\(x^3+12x^2+48x=-64\)
\(\Leftrightarrow x^3+12x^2+48x+64=0\)
\(\Leftrightarrow\left(x+4\right)^3=0\)
\(\Leftrightarrow x=-4\)
Vậy \(S=\left\{-4\right\}\)
