\(5.3^x=15.3^5\)

\(5.3^x=5.3.3^5\)

\(5.3^x=5.3^6\)

\(3^x=3^6\)

\(x=6\)

`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 ...

`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`

giúp mình với

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

\(\left(1+2+3+...+2017\right)\times\left(1717\times18-1818\times17\right)\\ =\left(1+2+3+...+2017\right)\times\left(17\times101\times18-18\times101\times17\right)\\ =\left(1+2+3+...+2017\right)\times0\\ =0\)