1/ ( x-3) 2=16
\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
2/ (3x-1)3=8
\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)
3/ (x-11)3=-27
\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)
phần 4 mình ko rõ đề
4) \(x^3-3x^2+3x-1=-64\)
\(\Rightarrow x^3-3x^2+3x+63=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(6x^2+18x\right)+\left(21x+63\right)=0\\ \Rightarrow x^2\left(x+3\right)+6x\left(x+3\right)+21\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x^2+6x+21\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x^2+6x+21=0\end{matrix}\right.\)
\(x+3=0\\ \Rightarrow x=-3\)
\(x^2+6x+21=0\\ \Rightarrow\left(x^2+6x+9\right)+12=0\\ \Rightarrow\left(x+3\right)^2+12=0\)
Vì \(\left(x+3\right)^2\ge0;12>0\Rightarrow\left(x+3\right)^2+12>0\Rightarrow x^2+6x+21vônghiệm\)
Vậy \(x=-3\)
