ƯC (42, 55, 91)=1

C

\(10\) chia hết cho 5 nên \(10^{2021}\) chia hết cho 5

Mà 8 ko chia hết cho 5

Nên \(10^{2021}+8\) không chia hết cho 5

D đúng

Đặt `A= 1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100)`

`3A=  3. (1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100))`

`3A=  1 + 1/3 + 1/(3^2) + ... + 1/(3^98) + 1/(3^99)`

`3A - A = (1 + 1/3 + 1/(3^2)+... + 1/(3^98) + 1/(3^99)) - (1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100))`

`2A = 1 - 1/(3^100)`

`A = (1 - 1/(3^100))/2`

Vậy: `1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100) = (1-1/(3^100))/2`

   A =         \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{99}}\) + \(\dfrac{1}{3^{100}}\)

3A =   1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{3^{98}}\) + \(\dfrac{1}{3^{99}}\)

3A - A = (1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ...+\(\dfrac{1}{3^{98}}\) + \(\dfrac{1}{3^{99}}\)) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{99}}\)+\(\dfrac{1}{3^{100}}\))

A.(3 - 1) = 1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{98}}\)+ \(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\) - ...- \(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3^{100}}\)

A x 2 =  (1 - \(\dfrac{1}{3^{100}}\)) + (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^{98}}\) - \(\dfrac{1}{3^{98}}\)) + (\(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3^{99}}\))

A x 2 = 1 - \(\dfrac{1}{3^{100}}\) + 0 + 0 + ..+ 0

A x 2 = 1 - \(\dfrac{1}{3^{100}}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{100}}\) 

25 = 5²

Ư(25) = {1; 5; 25}

Vì 3311 = 11 . 301 nên 3311 có ước là 11 và 301. Vậy 3311 là một hợp số.

Ta có: `3311 = 11 .301`

`->` Số `3311` là hợp số.

hợp số bạn nhé

`326 + (153 - x) = 403`

`=> 153 - x = 403 - 326`

`=> 153-x=77`

`=> x=153-77`

`=>x=76`

Vậy: `x=76`

326 + (153 - x) = 403

=> 153 - x = 403 - 326

=> 153-x=77

=> x=153-77

=>x=76

Vậy: x=76

`x` \(\in B\left(11\right),10< x< 40\)

\(\Rightarrow B\left(11\right)=\left\{0;11;22;33;44;...\right\}\)

Mà \(10< x< 40\)

\(\Rightarrow x\in\left\{11;22;33\right\}\)

x = 11;22;33

\(a,3^{x-2}\cdot2+15=33\)

\(\Rightarrow3^{x-2}\cdot2=33-15\)

\(\Rightarrow3^{x-2}\cdot2=18\)

\(\Rightarrow3^{x-2}=18:2\)

\(\Rightarrow3^{x-2}=9\)

\(\Rightarrow3^{x-2}=3^2\)

\(\Rightarrow x-2=2\)

` x = 2 +2 `

` x = 4`

Vậy `x = 4`

\(b,x⋮20,x⋮35\) và \(400< x< 500\)

Ta có :

$x \vdots 20 , x \vdots 35$ và `400 < x < 500`

$\Rightarrow x \in BC(20;35)$ $(ĐK : 400 <  x < 500)$

Ta có :

`20 = 2^2 `\(\cdot\)`5`

`35 = 5`\(\cdot\)`7` 

$\Rightarrow BCNN(20;35) = 2^2$\(\cdot\)$5$\(\cdot7\) $=140$

$B(140) = {0;140;280;420;560;...}$

Mà $400 < x < 500$

 $\Rightarrow x = 420$ 

   a; 3\(^{x-2}\).2 + 15 = 33

                   3\(^{x-2}\) = 33 - 15

                   3\(^{x-2}\).2 = 18

                   3\(^{x-2}\)    = 18 : 2

                   3\(^{x-2}\)    = 9

                  3\(^{x-2}\)     = 3²

                   \(x-2\)  = 2

                   \(x=2+2\)

                    \(x=4\)

              Vậy \(x=4\)

b; \(x⋮\) 20; \(x⋮\) 35 ⇒ \(x\) \(\in\) BC(20; 35) 

     20 = 2².5; 35 = 5.7 BCNN(20; 35) = 2².5.7 = 140

    \(x\) \(\in\) B(140) = {0; 140; 280; 420; 560; 840;..}

Vì 400 < \(x\) < 500 nên \(x\) = 420

Vậy \(x=420\) 

`a, 36 + 75 + 64 + 25 + 99`

`= (36 + 64) + (75 + 25) + 99`

`= 100 + 100 + 99`

`= 200 + 99`

`= 299`

`b, 2024^0 + 3 * [5^2 * 10 - (23 - 13)^2]`

`= 1 + 3*[5^2 * 10 - 10^2]`

`= 1 + 3*[25 * 10 - 100]`

`= 1 + 3*[250 - 100]`

`= 1 + 3* 150`

`= 1 + 450`

`= 451`

\(c,4^2\cdot65+35\cdot4^2-1500\)

`= 4^2 * (65 + 35) - 1500`

`= 16 * 100 - 1500`

`= 1600 - 1500`

`= 100`

a; 36 + 75  + 64 + 25+ 99

= (36 + 64) + (75 + 25) + 99

= 100 + 100 + 99

= 200 + 99

= 299

b;2024⁰ + 3.[5².10 - (23 - 13)²]

  1 + 3.[25.10 - 10²] 

= 1 + 3.[250 - 100]

= 1 +3.150

= 1 +450

= 451 

c; 4².65 + 35.4² - 1500

= 4².(65 + 35) - 1500

= 4².100 - 15.100

= 100.(4² - 15)

= 100.(16 - 15)

= 100.1

= 100