Vì (2x-1)^6=(2x-1)^8
(2x-1)^8-(2x-1)^6=0
(2x-1)^6[(2x-1)^2-1)]=0
th1 (2x-1)^6 suy ra 2x-1=0 suy ra x=1/2
th2 (2x-1)^2-1=0
(2x-1)^2=1
suy ra 2x-1 bằng 1;-1
th1 2x-1=1 suy ra x=1
2x-1=-1 suy ra x=0
Answer:
A, \(3^{x-1}+5.3^{x-1}=162\)
\(\Rightarrow3^{x-1}.\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
B, \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Rightarrow\left(2x-1\right)^6.[1-\left(2x-1\right)^2]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
C, Đề có chính xác không bạn nhỉ?
