Ta có: 

`54 = 2.3^3`

`40 = 2^3 . 5`

`=> BCN``N(40;54) = 2^3 . 3^3 . 5 = 1080`

`M = 3 + 3^2 + 3^3 + ... + 3^2024`

`3M = 3 . (3 + 3^2 + 3^3 + ... + 3^2024)`

`3M = 3^2 + 3^3  + 3^4 + ... + 3^2025`

`3M - M = (3^2 + 3^3 + 3^4 + ... + 3^2025) - (3 + 3^2 + 3^3 + ... + 3^2024)`

`2M = 3^2025 - 3`

`M = (3^2025 - 3)/2`

`M = 3 + 3^2 + 3^3 + ... + 3^2024`

`3M = 3 . (3 + 3^2 + 3^3 + ... + 3^2024)`

`3M = 3^2 + 3^3 + 3^4 +... + 3^2025`

`3M - M = ( 3^2 + 3^3 + 3^4 +... + 3^2025) - (3 + 3^2 + 3^3 + ... + 3^2024)`

`2M = 3^2025 - 3`

`M = (3^2025 - 3)/2`

\(\left(84,6-2\times x\right)\div3,02=5,1\)

\(84,6-2\times x=5,1\times3,02\)

\(84,6-2\times x=15,402\)

\(2\times x=84,6-15,402\)

\(2\times x=69,198\)

\(x=69,198\div2\)

\(x=34,599\).

~ HOK TỐT ~

84,6-2*x=5,1*3,02

84,6-2*x=15,402

        2*x=84,6-15,402

        2*x=  69,198

          x=69,198/2

          x=34,599

`4.2^x +` $2^{x+3} = 48$

$\Rightarrow 4 .$ `2^x + 2^x . 8` $= 48$

$\Rightarrow (4+8)$`.2^x``=48`

$\Rightarrow 12 . $`2^x` `=48`

$\Rightarrow $`2^x = 48 : 12`

$\Rightarrow $`2^x = 4`

$\Rightarrow$`2^x = 2^2`

$\Rightarrow$ `x = 2`

Vậy `x = 2`

`(-38) . 63 + (-37) . 38`

`= (-38) . 63 + 37 . (-38)`

`= (-38) . (63+37)`

`= (-38) . 100`

`= -3800`

 =(-38) + 38 . 63 .(-37) 

= 0 . 2431

= 0

`((-14).11+14.2 )/( 11.21-7.22)`

`= (14. (-11)+14.2 )/( 11.21-7.2 .11)`

`= (14. (-11)+14.2 )/( 11.21-14 .11)`

`= (14. (-11+2) )/( 11.(21-14))`

`= (14. (-9) )/( 11.7)`

`= (2. (-9) )/11`

`=  - 18/11`

Điểm anh của tôi được 1,75

`18 : 3^2 + 5 . 2^3`

`= 18 : 9 + 5 . 8`

`= 2 + 40 `

`= 42`

Chỗ 4 mũ 2/3.5 x ... x 59 mũ 2/58.60 nha

a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)

                                                                                   \(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)

=> đpcm

Study well ! >_<

Nếu A chia hết cho B thì A là bội của B

`3n - 1 vdots n`

`=> 3n - 1 - n - n - n vdots  n`

``=> 3n - 1 - 3n vdots n`

`=> -1 vdots n`

`=> n in Ư(-1) = {-1;1}`

Vậy ...

b; (3n + 1) ⋮ (2n - 1) 

    2(3n + 1)⋮ (2n  -1)

    (6n + 2) ⋮ (2n  - 1)

  [3.(2n - 1) + 5] ⋮ (2n - 1)

        5 ⋮ (2n  -1)

    (2n  - 1) ϵ Ư(5) = {-5; -1; 1; 5}

     Lập bảng ta có:

Theo bảng trên ta có: n ϵ {-2; 0; 1; 3}

Vậy các giá trị nguyên của n là: n ϵ {-2; 0; 1; 3}