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)=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 \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

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

minh ko bit

(2x-1)⁶ = (2x-1)⁸

=> 2x-1 \(\in\){-1; 0; 1}

=> 2x \(\in\){0; 1; 2}

=> x \(\in\){0; 1/2; 1}

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

b)2^x.(1+2³⁾=144

2^{x .}(1+8)=144

2^{x .} 9 =144

2^x     =144:9

2^x     =16

2^{x       =}2⁴

x      =4

câu a ko chả lời được đâu vì

2x - 1 giống nhau thì coi như là bằng

còn mủ 6 và 8 sao bằng được

chỉ có sai đề

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

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

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

\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0,5\\x\in\left\{0;1\right\}\end{cases}\Rightarrow}x\in\left\{0;0,5;1\right\}}\)

\(b,2^x+2^{x+3}=144\)

\(\Rightarrow2^x.1+2^x.8=144\)

\(\Rightarrow2^x.9=144\)

\(\Rightarrow2^x=16=2^4\Rightarrow x=4\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=2\\2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)

\(b,5^x+5^{x+1}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x.\left(1+5^2\right)\)\(=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=650\div26\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

\(c,3^{x-1}+5.3^{x-1}=162\)

\(\Leftrightarrow3^{x-1}.\left(1+5\right)=162\)

\(\Leftrightarrow3^{x-1}.6=162\)

\(\Leftrightarrow3^{x-1}=162\div6\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=3+1\)

\(\Leftrightarrow x=4\)

thanks nhìu

(2x-1)² = (2x-1)⁶

(2x-1)² - (2x-1)⁶ = 0

(2x-1)²  x (1-(2x-1)⁴) = 0

=> 2x-1 = 0 <=> x=1/2

=> 2x-1=1 <=> x=1

vội cũng phải cho mik nha

cái này thì mình ko biết rõ cách làm kết quả là 0 hoặc 1

Ns thiệt nha thik ai chứ thik CHELSEA mik hơi bị hate

Nhưng mà mik agree help bn lun

Biểu thức căn 2 =nhau khi và chỉ khi

1^2=1^6

hoặc

0^2=0^6

và cả trường hợp âm nữa nhé

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

\(\Rightarrow2x-1=0\Rightarrow x=0,5\)

\(\Rightarrow2x-1=1\Rightarrow x=1\)

\(\Rightarrow2x-1=-1\Rightarrow x=0\)

\(\Rightarrow x\in\left\{0,5,0;1\right\}\)