a)
\(\left(3x-2\right)^3=1\)
\(\Rightarrow\left(3x-2\right)^3=1^3\)
\(\Rightarrow3x-2=1\)
\(\Rightarrow x=1\)
b)
\(\left(x-3\right)^6=\left(x-3\right)^3\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-3\right)^6:\left(x-3\right)^3=1\\\left(x-3\right)^6-\left(x-3\right)^3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-3\right)^3=1\\\left(x-3\right)^3\left[\left(x-3\right)^3-1\right]=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=1\\\left[\begin{array}{nghiempt}x-3=0\\\left(x-3\right)^3-1=0\end{array}\right.\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\\left[\begin{array}{nghiempt}x=3\\\left(x-3\right)^3=1\end{array}\right.\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\end{array}\right.\)
Vậy \(x\in\left\{3;4\right\}\)
\(\left(3x-2\right)^3=1\)
\(\Rightarrow27x^3-8=1\)
\(\Rightarrow27x^3=9\)
\(\Rightarrow x^3=\frac{9}{27}\)
\(x^3=\frac{1}{3}\)
\(\left(x-3\right)^6=\left(x-3\right)^3\)
\(x^6-3^6=x^3-3^3\)
\(-3^3=-x^3\)
\(\Rightarrow-3=-x\)
\(\Rightarrow x=3\)
a) \(\left(3x-2\right)^3=1\)
\(\Rightarrow3x-2=1\)
\(3x=1+2\)
\(3x=3\)
\(\Rightarrow x=1\)
