ket qua x=1

Các giá trị x thu được là \(0,1,\frac{1}{2}\)

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

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

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

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

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

Còn bn tự giải nhé

Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

giống cái kia thôi bn

Mik làm rồi mà

Mà cái bn Nguyễn Duy Đạt gì đó làm thiếu 1 trường hợp

Mà bn vẫn kik hở

Sao zzzzz??????

chắc nhầm ! Xin lỗi ! Xin lỗi !

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

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

''...'' vế sau á :), ko chắc 

(2x-1)⁶ = (2x-1)⁸

=> 2x-1 \(\in\){-1; 0; 1}

=> 2x \(\in\){0; 1; 2}

=> x \(\in\){0; 1/2; 1}

.............................................................................................................................................................................................????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????.....................................................................................................................................................................?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

minh ko bit

(2x-1)^6=(2x-1)^8'

=> (2x-1)=1 hoặc (2x-1)=0

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

(2x-1)=0 =>2x=1 =>x=0,5

mik nghĩ x=0,5 k mik nha

 chs bạn học giỏi^_^

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

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

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

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

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

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

\(\left(2x-1\right)^6.\left[4x\left(1-x\right)\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\4x\left(1-x\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\4x=0hoac1-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0hoacx=1\end{cases}}\)

vay \(x=\frac{1}{2}\)hoac \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(\frac{\left(2-3x\right)^4}{\left(3x-2\right)^5}=\frac{-\left(3x-2\right)^4}{\left(3x-2\right)^5}=\frac{-1}{3x-2}\)