a) (x - 2)8 = (x - 2)6
=> (x - 2)8 - (x - 2)6 = 0
=> (x - 2)6.[(x - 2)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\\left(x-2\right)^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\x-2=1\\x-2=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=3\\x=1\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=2\\x=3\\x=1\end{array}\right.\)
b) \(\frac{3}{4}x-\frac{1}{2}\left(x-2\right)+\frac{2}{3}=0\)
\(\Rightarrow\frac{3}{4}x-\frac{1}{2}x+1+\frac{2}{3}=0\)
\(\Rightarrow\frac{1}{4}x+\frac{5}{3}=0\)
\(\Rightarrow\frac{1}{4}x=0-\frac{5}{3}=\frac{-5}{3}\)
\(\Rightarrow x=\frac{-5}{3}:\frac{1}{4}\)
\(\Rightarrow x=\frac{-5}{3}.4=\frac{-20}{3}\)
Vậy \(x=\frac{-20}{3}\)
