`8/9 x - 2/3 = 1/3x + 1 1/3`

`8/9 x - 2/3 = 1/3 x + 4/3`

`8/9 x - 1/3 x = 2/3 + 4/3`

`8/9 x - 1/3 x = 6/3`

`8/9 x - 1/3 x = 2`

`(8/9-1/3)x = 2`

`(8/9 - 3/9)x =2`

`5/9 x = 2`

`x=2 : 5/9`

`x= 2 . 9/5`

`x= 18/5`

Vậy`x=18/5`

8/9*x - 2/3 = 1/3*x + 1/1/3

=> 8/9*x - 1/3*x = 2/3 + 1/1/3

=> x*(8/9 - 1/3) = 2/3 + 1*3/1

=> x*5/9 = 11/3

=> x = 11/3 / 5/9 = 11/3 * 9/5 = 11/1 * 3/5

=> x = 33/5

Vậy x = 33/5

a) 2³ * 128 * 64 = 2³ * 2⁷ * 2⁶= 2¹⁶

b) 3⁴ * 243 * 81 = 3⁴ * 3⁵ * 3^{3 =} 3¹²

c) 5³ * 125 * 625 = 5³ * 5³ * 5⁴ = 5¹⁰

d) 7² * 7³ * 49 = 7² * 7³ * 7² = 7⁸

`49 : (2^x -9) = 4^2 -3^2`

`49 : (2^x -9) =16-9`

`49 : (2^x -9) =7`

`2^x - 9 =49:7`

`2^x -9=7`

`2^x = 7+9`

`2^x = 16`

`=> 2^x = 2^4`

`=> x=4`

\(3^{x+1}.5-10:2+5=15\)

`3^(x+1) . 5 - 5 + 5 = 15`

`3^(x+1) .5 - 5=15-5`

`3^(x+1).5 -5 = 10`

`3^(x+1).5=10+5`

`3^(x+1).5=10=5`

`3^(x+1) = 15 : 5`

`3^(x+1)=3`

`3^(x+1)=3^1`

`x+1=1`

`x=1-1`

`x=0`

Vậy `x=0`

3 đúng không?

\(A=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)\) chia hết cho \(3\).

\(\left(7^8+7^9+7^{10}\right):7^7\\ =\dfrac{7^8}{7^7}+\dfrac{7^9}{7^7}+\dfrac{7^{10}}{7^7}\\ =7^1+7^2+7^3=399=57\times7⋮57\)

không chia hết cho 57(nghĩa là có dư khi chia cho 57)

7^20+7^19-7^18                                                                                        =7^18(7^2+7-1)
=7^18(49+7-1)
=7^18x55
=7^18x11x5 chia hết cho 11                                                                               Vậy......

\(\dfrac{6n+3}{3n+6}=\dfrac{6n+12-9}{3n+6}=\dfrac{2\left(3n+6\right)-9}{3n+6}=2-\dfrac{9}{3n+6}\)                           ⇒3n+6 ϵ Ư(9)={1,3,9,-1,-3,-9}                                                                 Theo bài ta có bảng sau:

⇒Theo bài n là số nguyên nên n ϵ {0,1,-1,-2}