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Điều kiện; n nguyên
Ta có: \(\left(5\text{​​}n-9\right)⋮n\)
Vì \(5n⋮n\) nên \(-9⋮n\)
\(\Rightarrow n\inƯ\left(-9\right)=\left\{\pm1,\pm3,\pm9\right\}\) 9thỏa mãn)
Vậy...

Bổ sung: `n` thuộc `Z`

Ta có: `5n-9` và `n` thuộc `Z; n ≠ 0`

`5n - 9 ⋮ n`

Do `n ⋮ n => 5n ⋮ n`

`=> 9 ⋮ n`

`=> n` thuộc `Ư(9) =` {`-9;-3;-1;1;3;9`} (Thỏa mãn)

Vậy ...

Đặt \(A=-2^{49}-2^{48}-...-2^1-1\)
\(\Rightarrow-A=2^{49}+2^{48}+...+2^1+1\\ \Rightarrow-2A=2^{50}+2^{49}+...+2^2+2^1\\ \Rightarrow-A-\left(-2A\right)=\left(2^{49}+2^{48}+...+2^1+1\right)-\left(2^{50}+2^{49}+...+2^2+2^1\right)\\ A=1-2^{50}\)
Thay vào \(2^{50}-2^{49}-2^{48}-...-2^1-1\) được:
\(2^{50}-2^{49}-2^{48}-...-2^1-1\\ =2^{50}+1-2^{50}\\ =1\)

`S = 2^50 -2^49 -2^48 -...-2^1 -1`

`2S = 2^51 - 2^50 - 2^49 - ... - 2^2 - 2`

`2S - S = (2^51 - 2^50 - 2^49 - ... - 2^2 - 2) - (2^50 -2^49 -2^48 -...-2^1 -1)`

`S =  2^51 - 2^50 - 2^49 - ... - 2^2 - 2 - 2^50 +2^49 +2^48 +...+2^1 +1`

`S = 2^51 - 2^50 - 2^50  + 1`

`S = 2^51 - (2^50 + 2^50)  + 1`

`S = 2^51 - 2.2^50   + 1`

`S = 2^51 - 2^51   + 1`

`S = 1`

a: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

=>x=3k; y=4k; z=5k

\(2x^2+2y^2+3z^2=-100\)

=>\(2\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2=-100\)

=>\(125k^2=-100\)

=>\(k^2=-\dfrac{4}{5}\)(vô lý)

vậy: \(\left(x;y;z\right)\in\varnothing\)

b: 2x=y/3=z/5

=>\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=>\(x=\dfrac{1}{2}k;y=3k;z=5k\)

\(x+y-\dfrac{z}{2}=-20\)

=>\(\dfrac{1}{2}k+3k-\dfrac{5k}{2}=-20\)

=>k=-20

=>\(x=\dfrac{1}{2}\cdot\left(-20\right)=-10;y=3\cdot\left(-20\right)=-60;z=5\cdot\left(-20\right)=-100\)

Ta có 

\(111=3.37\Rightarrow n+2=\left\{3;37;111\right\}\Rightarrow n=\left\{1;35;109\right\}\)

\(\Rightarrow n-2=\left\{-1;33;107\right\}\)

Ta thấy n-2 =33 là bội của 11

=> n=35

giúp mình với

`S = 3^100 -3^99 +3^98 -3^97 +...+3^2 -3 +1`

`3S = 3^101 - 3^100 +3^99- 3^98+...+ 3^3 -3^2 +3`

`S + 3S = (3^100 -3^99 +3^98 -3^97 +...+3^2 -3 +1) + (3^101 - 3^100 +3^99- 3^98+...+ 3^3 -3^2 +3)`

`4S = 3^101 + (3^100 - 3^100) + (3^99 - 3^99) + ... + (3 - 3) + 1`

`4S = 3^101 + 1`

`S = (3^101 + 1)/4`

`H=1+2.6+3.6^2+4.6^3+...+100.6^99`

`6H = 6+2.6^2+3.6^3+4.6^4+...+100.6^100`

`6H - H = (6+2.6^2+3.6^3+4.6^4+...+100.6^100)-(1+2.6+3.6^2+4.6^3+...+100.6^99)`

`5H = (6 - 2.6) + (2.6^2 - 3.6^2) + (3.6^3  - 4.6^3) + ... + (99. 6^99 - 100.6^99) + 100.6^100 - 1`

`5H = 100.6^100 - 1 + (-6) + (-6^2) + (-6^3) + ... + (-6^99)`

`5H = 100.6^100 - 1 - (6+6^2+6^3 + ... + 6^99)`

Đặt `S = 6+6^2+6^3 + ... + 6^99`

`6S = 6^2+6^3+6^4 + ... + 6^100`

`6S - S = (6^2+6^3+6^4 + ... + 6^100) - ( 6+6^2+6^3 + ... + 6^99)`

`5S = 6^100 - 6`

`S = ( 6^100 - 6)/5`

Khi đó: `5H = 100.6^100 - 1 - S`

`5H = 100.6^100 - 1 - ( 6^100 - 6)/5`

`5H = (500.6^100)/5 - 5/5 - ( 6^100 - 6)/5`

`5H =  (500.6^100 - 5 - 6^100 + 6)/5`

`H = (499 . 6^100 + 1)/5`

Vậy ...

Ta có: \(\widehat{xOt}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}=180^0-70^0=110^0\)

Ta có: \(\widehat{xOt}=\widehat{yOz}\)(hai góc đối đỉnh)

mà \(\widehat{xOt}=110^0\)

nên \(\widehat{yOz}=110^0\)

Ta có: \(\widehat{yOz}+\widehat{zOt}=180^0\)(hai góc kề bù)

=>\(\widehat{zOt}=180^0-110^0=70^0\)

Do `Oz` đối tia `Ox`

=> \(\widehat{xOz}=180^o\)

Mà \(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\) (Vì `Oy` nằm giữa `Oz` và `Ox`)

=> \(70^o+\widehat{yOz}=180^o\)

=> \(\widehat{yOz}=180^o-70^o\)

=> \(\widehat{yOz}=110^o\)

Lại có: \(\left\{{}\begin{matrix}\widehat{xOy}=\widehat{tOz}\\\widehat{yOz}=\widehat{xOt}\end{matrix}\right.\) (Các cặp góc đối đỉnh)

=> \(\left\{{}\begin{matrix}\widehat{zOt}=70^o\\\widehat{xOt}=110^o\end{matrix}\right.\)

Vậy ...