Sửa đề: 

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

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

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

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

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

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

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

`=> 2A = 1 - 1/(3^99)`

`=> A = (1 - 1/(3^99))/2`

Khi đó: `2S = 1 - 100/(3^100) + (1 - 1/(3^99))/2`

`S = 1/2 - 100/(2.3^100) + (1 - 1/(3^99))/4`

Ta có: `{(1/2 - 100/(2.3^100) < 1/2),((1 - 1/(3^99))/4 < 1/4):}`

`=>  1/2 - 100/(2.3^100) + (1 - 1/(3^99))/4 < 1/2 + 1/4 = 3/4`

Hay `S < 3/4 (đpcm)`

3|4lớn hơn

tk nhé

Ông An cao 180 cm, vòng bụng 108 cm.

Ông Chung cao 160 cm, vòng bụng 70 cm.

`1/4 . 2/6 . 3/8 . 4/10 . ... . 31/64 = 2^x`

`=> 1/(2.2) . 2/(2.3) . 3/(2.4) . 4/(2.5) . ... . 31/(32.2) = 2^x`

Số phân số có trong dãy là: `(31 - 1) : 1 + 1 = 31` (phân số) 

`=> (1.2.3.4...31)/(2^31 . 2 . 3 . 4 . 5 ... 31.32) = 2^x`

`=> 1/(2^31 . 32) = 2^x`

`=> 1/(2^31 . 2^5) = 2^x`

`=> 1/(2^(31+5)) = 2^x`

`=> 1/(2^36) = 2^x`

`=> 2^(-36) = 2^x`

`=> x = -36`

Vậy `x = -36`

`(1/3)^(2x - 1) = 3^5`

`=> (3^(-1))^(2x - 1) = 3^5`

`=> 3^(-1.(2x-1)) =  3^5`

`=> 3^(1-2x) = 3^5`

`=> 1 - 2x = 5`

`=> 2x = 1 - 5`

`=> 2x = -4`

`=> x = -2`

Vậy `x = -2`

\(-x+5=3x+\left(-11\right)\)

\(-x-3x=-5+\left(-11\right)\)

\(-4x=-16\)

\(x=\left(-16\right):\left(-4\right)\)

\(x=4\)

Đa dạng cách nhé :Đ

`-x + 5 = 3x + (-11)`

`=> 3x + x = 5 + 11`

`=> 4x = 16`

`=> x = 16 : 4`

`=> x = 4`

Đề là gì bạn?

\(n^2+n-7=\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để: \(\left(n^2+n-7\right)⋮\left(n-2\right)\Rightarrow\left[\left(n-2\right)\left(n+3\right)-1\right]⋮\left(n-2\right)\\ \Rightarrow1⋮\left(n-2\right)\) (Vì: \(\left(n-2\right)\left(n+3\right)⋮\left(n-2\right)\forall n\inℤ\) )

\(\Rightarrow n-2\in\left\{1;-1\right\}\Rightarrow n\in\left\{3;1\right\}\)

Ta có:

\(n^2+n-7\\ =\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để `n^2+n-7` chia hết cho n - 2 thì: 

1 ⋮ n - 2

=> n - 2 ∈ Ư(1) = {1; -1}

=> n ∈ {3; 1} 

Chọn B

B

Cho đa thức: \(-8x^2+8x+8=0\)

\(\Rightarrow-8\left(x^2-x-1\right)=0\\ \Rightarrow x^2-x-1=0\\ \Rightarrow\left(x^2-x-\dfrac{1}{4}\right)-\dfrac{3}{4}=0\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}\\ \Rightarrow x-\dfrac{1}{2}=\pm\dfrac{\sqrt{3}}{2}\\ \Rightarrow x=\dfrac{1\pm\sqrt{3}}{2}\)