Question

Tìm x biết

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

Answer 1

x=1 và 0  

ms thỏa mản đề ra

=)))))))))))))))

Answer 2

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

Answer 3

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

Answer 4

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

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

\(\Rightarrow1=\left(2x-1\right)^2\)

\(\Rightarrow1^2=\left(2x-1\right)^2\)

\(\Rightarrow2x-1=1\)

\(\Rightarrow2x=1+1\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=2\div2\)

\(\Rightarrow x=1\)

vậy \(x=1\)

Answer 5

x = 0 vẫn cho kết quả đúng nhé bạn :))

Answer 6

mình không biết làm.

Answer 7

http://123link.pw/DCUeSUw

Answer 8

pt<=> (2x-1)^6 . [ (2x-1)^2 -1 ]=0

<=>    (2x-1)^6 . [ 4x^2-4x ]=0

<=> x . (2x-1)^6 . (x-1) = 0

=> x = 0 hoặc x = 1/2 hoặc x= 1

Answer 9

(2x-1)⁸-(2x-1)⁶=0

(2x-1)⁶  [(2x-1)²-1]=0

=>x=1/2 hoặc x=1

Answer 10

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

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

=>\(\left(2x-1\right)^6\).1-\(\left(2x-1\right)^6\).\(\left(2x-1\right)^2\)=0

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

=>\(\left(2x-1\right)^6\)=0 hoặc \(1-\left(2x-1\right)^2\)=0

=>2x-1=0 hoặc 2x-1=1

=> x=\(\frac{1}{2}\)hoặc x=1

vậy x=\(\frac{1}{2}\)hoặc x=1