Question

Tìm x 

\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

Answer 1

x=1 nh bạn

Answer 2

Ta có \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

\(\Rightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)^2-1\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\\left(2x-1\right)^2=1\end{cases}}}\)

Ta có \(\left(2x-1\right)^2=1\Rightarrow\hept{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)

Answer 3

=> (2x - 1)⁸ - (2x - 1)⁶  = 0

=> (2x - 1)⁶ .[(2x-1)²-1] = 0

=> (2x-1)⁶.4x.(x-1)=0

=>\(\hept{\begin{cases}2x-1=0\\4x=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)

Vậy x=1/2 hoặc x=0 hoặc x=1

Answer 4

Vì (2x - 1) ^6 = (2x - 1) ^8

 TH1:

=>(2x - 1) ^6=0 

(2x - 1) ^8=0

Nhưng không có kết quả nào tướng ứng nên TH này loại

TH2:

(2x - 1) ^6=1=>2x - 1 = 1 =>x=1

(2x - 1) ^8=1=>2x - 1 = 1 =>x=1

Nên => x=1

Answer 5

Ta có \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
         \(\Rightarrow\left(2x-1\right)^8:\left(2x-1\right)^6=1\)
          \(\Rightarrow\left(2x-1\right)^2=1\)
            \(\Rightarrow\left(2x-1\right)^2=\left(\pm1\right)\)
Nếu \(\hept{\begin{cases}\left(2x-1\right)^2=1\Rightarrow2x-1=1\Rightarrow2x=2\Rightarrow x=1\\\left(2x-1\right)^2=-1\left(loại\right)\end{cases}}\)
                                           Vậy x = 1