Question

Tìm \(x\in Q\), biết:

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

Answer 1

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

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

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

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

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

Đến đây bạn tự giải nha.

x=1/2 hoặc x=1 hoặc x=0

\(\Rightarrow\orbr{\begin{cases}2x=1\\\orbr{\begin{cases}\\2x-1=-1\end{cases}}2x-1=1\end{cases}}\)

Answer 2

suy ra 2x -1=0

2x=1

x=1:2

x=\(\frac{1}{2}\)

Answer 3

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

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

2x - 1 = 0 => x = 1/2(2x - 1)² = 1 => 2x - 1 = -1 => x = 0 hoặc 2x - 1 = 1 => x = 1.

PT có 3 nghiệm: x = 0; 1/2; 1.

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

Answer 4

\(\frac{1}{2}\)