(3x - 7)8 = (3x - 7)6
=> (3x - 7)8 - (3x - 7)6 = 0
=> (3x - 7)6.[(3x - 7)2 - 1) = 0
=> \(\orbr{\begin{cases}\left(3x-7\right)^6=0\\\left(3x-7\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}3x=7\\3x-7\in\left\{1;-1\right\}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\3x\in\left\{8;6\right\}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}\)
\(\left(3x-7\right)^8=\left(3x-7\right)^6\)
TH1\(\left(3x-7\right)^2=\left(3x-7\right)^6:\left(3x-7\right)^6\)
\(\left(3x-7\right)^2=1\)
\(\Rightarrow3x-7=1\)
\(3x=8\)
\(x=\frac{8}{3}\)
TH2 \(3x-7=0\)
\(3x=7\)
\(x=\frac{7}{3}\)
Vậy \(x=\frac{8}{3};x=\frac{7}{3}\)
(3x - 7)8 = (3x - 7)6
=> (3x - 7)8 - (3x - 7)6 = 0
=> (3x - 7)6.[(3x - 7)2 - 1) = 0
=> $$=> $$=> $$
=> $$=> $$
Vậy $x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}$
