ta có : \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)(=)\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)(=)\(\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)(=)\(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left[\left(4x-1\right)^{10}-1\right]=0\end{cases}}\)(=)\(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)(=)\(\orbr{\begin{cases}4x=1\\\begin{cases}4x-1=1\\4x-1=-1\end{cases}\end{cases}}\)(=)\(\orbr{\begin{cases}x=\frac{1}{3}\\\begin{cases}4x=2\\4x=0\end{cases}\end{cases}}\)\(\orbr{\begin{cases}x=\frac{1}{4}\\\begin{cases}x=\frac{1}{2}\\x=0\end{cases}\end{cases}}\)

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\Leftrightarrow\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=-1;1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{1}{4};0;\frac{1}{2}\)

P/s : phần \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)   thay dấu \(\hept{\begin{cases}\\\\\end{cases}}\) thành dấu \(\orbr{\begin{cases}\\\end{cases}}\) nhé!
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\end{cases}}\)

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\Leftrightarrow\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x-1=\pm1\end{cases}}\)

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

Vậy x = 1/4 hoặc 1/2 hoặc 0 

số mủ của 1 dù lớn đến bao hiêu thì cũng bằng 1 suy ra x=1 vậy thay x=1 vào biễu thức là ra

Ta có:

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

Xét \(4x-1=0\)

\(\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\), thỏa mãn

Xét \(4x-1\ne0\)

\(\Rightarrow\left(4x-1\right)^{30}:\left(4x-1\right)^{20}=1\)

\(\Rightarrow\left(4x-1\right)^{10}=1\Rightarrow\left[{}\begin{matrix}4x-1=1\Rightarrow x=\frac{1}{2}\\4x-1=-1\Rightarrow x=0\end{matrix}\right.\)

Vậy....

Bài 1 :

\(C=\frac{1}{\left|x-2\right|+3}\)

\(C\le\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

Bài 2 :

a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)

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

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2\cdot3^{x-405}=3^{x-1}\)

\(2=3^{x-1}:3^{x-405}\)

\(2=3^{x-1-x+405}\)

\(2=3^{404}\)( vô lí )

=> x thuộc rỗng

c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)

\(\frac{27^{2x}}{81}=9^4\)

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

\(\frac{3^{6x}}{3^4}=3^8\)

\(3^{6x-4}=3^8\)

\(\Rightarrow6x-4=8\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)

a. ta có \(\left(4x+4\right)\left(3y+1\right)=20\Leftrightarrow\left(x+1\right)\left(3y+1\right)=5\)

Vậy \(\hept{\begin{cases}x+1=1\\3y+1=5\end{cases}}\text{ hoặc }\hept{\begin{cases}x+1=5\\3y+1=1\end{cases}}\)

Vì x,y là các số tự nhiên nên ta giải ra \(\hept{\begin{cases}x=4\\y=0\end{cases}}\)

b. ta có 2y+1 là số lẻ và là ước của 30 nên 

\(2y+1\in\left\{1,3,5,15\right\}\)

Với \(2y+1=1\Rightarrow x-1=30\Rightarrow\hept{\begin{cases}y=0\\x=31\end{cases}}\)

Với \(2y+1=3\Rightarrow x-1=10\Rightarrow\hept{\begin{cases}y=1\\x=11\end{cases}}\)

Với \(2y+1=5\Rightarrow x-1=6\Rightarrow\hept{\begin{cases}y=2\\x=7\end{cases}}\) Với \(2y+1=15\Rightarrow x-1=2\Rightarrow\hept{\begin{cases}y=7\\x=3\end{cases}}\)

\(\Rightarrow124-\left(20-4x\right)=120\\ \Rightarrow20-4x=24\\ \Rightarrow4x=-4\\ \Rightarrow x=-1\)

[ 124 - ( 20 - 4x ) ] : 30 = 4 

=> [ 124 - ( 20 - 4x ) ] : 30 = 4

=> 124 - ( 20 - 4x ) = 120

=> 20 - 4x = 124 - 120

=> 20 - 4x = 4

=>4x = 20 - 4

=>4x = 16

=> x = 16 : 4

=> x = 4

Vậy x = 4 .

\(\left[124-\left(20-4x\right)\right]:30=4\)

\(\Leftrightarrow20-4x=4\)

hay x=4

a)Ta co: x+20 la boi cua x+2

=>(x+20)chia het cho(x+2)

=>(x+2)+18chia het cho (x+2)

=>18 chia het cho (x+2)

=>(x+2) thuoc Ư(18)

Mà Ư(2)= 1;2;3;6;9;18

ta có bảng sau:

Vậy x = 0;1;4;7;16.

Nếu đúng nhớ tặng mình và đúng nhé!

Thank you!

x = 0,1,4,7,16 nha

4x + 3x = 30 – 20 : 10

7x = 30 - 2

7x = 28

x = 28 : 7

x = 4