a 54

b 41

k cho minh nhe ket ban khong

\(a,123+x=23\times3\)

    \(123+x=69\)

                 \(x=69-123\)

                 \(x=-54\)

a,123+x=69                                 

x=69-123

x=-54

B,\(2x^3=2^3\)

Suy ra:

x=2:2

x=1

\(a,\left(x-36\right):\left(2\cdot3^2\right)=2^3\cdot3\\ \Leftrightarrow x-36=432\\ x=468\\ b,2^x=32\\ \Leftrightarrow x=5\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x^3=3^3\\ \Leftrightarrow x=3\\ d,1579+\left(625-x\right)=2023\\ \Leftrightarrow x=1579+625-2023\\ \Leftrightarrow x=181\)

A. \(\left(x-36\right):\left(2.3^2\right)=2^3.3\)

\(\left(x-36\right):\left(2.9\right)=8.3\)

\(\left(x-36\right):18=24\)

\(x-36=24.18\)

\(x-36=432\)

\(x=432+36\)

\(x=468\)

B. \(2^x=32\)

\(2^x=2^5\)

\(x=5\)

C. \(x^3=27\)

\(x^3=3^3\)

\(x=3\)

D. \(1579+\left(625-x\right)=2023\)

\(625-x=2023-1579\)

\(625-x=444\)

\(x=625-444\)

\(x=181\)

a) \(\left(x-36\right)\div\left(2\times3^2\right)=2^3\times3\)

\(\left(x-36\right)\div\left(2\times9\right)=8\times3\)

\(\left(x-36\right)\div18=24\)

\(x-36=432\)

\(x=468\)

b) \(2^x=32\)

\(2^x=2^5\)

\(x=5\)

c) \(x^3=27\)

\(x^3=3^3\)

\(x=3\)

d) \(1579+\left(625-x\right)=2023\)

\(\left(625-x\right)=2023-1579\)

\(625-x=444\)

\(x=625-444\)

\(x=181\)

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

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

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

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

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

Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

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

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

a. (2x-1)⁴=81

=>(2x-1)⁴=3⁴

=>2x-1=3

=>2x=3+1

=>2x=4

=>x=4:2

=>x=2

b.(x-1)⁵=-32

=>(x-1)⁵=(-2)⁵

=>x-1=-2

=>x=-2+1

=>x=-1

c.(2x-1)⁶=(2x-1)⁸

mà chỉ có: (-1)⁶=(-1)⁸; 0⁶=0⁸; 1⁶=1⁸

=> để (2x-1) \(\in\){-1;0;1} thì x \(\in\){0; 1/2; 1}

(2x-1)^4=3^4

=>2x-1=3

=>2x=4

=>x=2

ai h minh minh h lai cho

a,  3 x + 1 : 3 4 = 81

3 x - 3 = 3 4

x – 3 = 4

x = 7

Vậy x = 7

b,  3 x + 3 . 3 x + 1 = 729

3 2 x + 4 = 3 6

2x + 4 = 6

x = 1

Vậy x = 1

c,  2 x + 3 . 2 x = 128

2 2 x + 3 = 2 7

2x + 3 = 7

x = 2

Vậy x = 2

d,  23 + 3 x = 5 6 : 5 3

23 + 3 x = 5 3

23 + 3x = 125

3x = 102

x = 34

Vậy x = 34

e,  2 x + 2 x + 4 = 272

2 x + 2 x . 2 4 = 272

2 x ( 1 + 2 4 ) = 272

2 x . 17 = 272

2 x = 16

2 x = 2 4

x = 4

Vậy x = 4

a,  2 3 x + 5 2 x = 2 5 2 + 2 3 - 33

8x+25x = 33

33x = 33

x = 1

b,  260 : x + 4 = 5 2 3 + 5 - 3 3 2 + 2 2

260:(x+4) = 5.13–3.13

x+4 = 260:26

x+4 = 10

x = 6

c,  720 : [ 41 - 2 x - 5 ] = 2 3 . 5

41–(2x–5) = 720:40

2x–5 = 41–18

2x = 28

x = 14

d,  3 2 - 2 x - 12 + 35 = 5 2 + 279 : 3 2

7(x–12)+35 = 56

7(x–12) = 21

x–12 = 3

x = 15

minh ko bit