\(2^{x+2}-2^x=96\)

\(2^x.\left(2^2-1\right)=96\)

\(2^x.3=96\)

\(2^x=96:3\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

A={1;3;2}

$9 . 2^{3x-1} = 36$

$\Rightarrow 2^{3x-1} = 36 : 9$

$\Rightarrow 2^{3x-1} = 4$

$\Rightarrow 2^{3x-1} = 2^{2}$

$\Rightarrow 3x - 1 = 2$

$ 3x = 2 + 1$

$ 3x = 3$

$ x = 3 : 3$

$ x = 1$

Vậy `x = 1`

\(3^{x-4}-17=2^6\)

\(\Rightarrow3^{x-4}-17=64\)

\(\Rightarrow3^{x-4}=64+17\)

\(\Rightarrow3^{x-4}=81\)

\(\Rightarrow3^{x-4}=3^4\)

$\Rightarrow x - 4 = 4$

$ x = 4 + 4$

$ x = 8$

Vậy `x = 8`

 x = 1

 x = 8

`105 +(3+x)^2=121`

`=> (3+x)^2=121-105`

`=> (3+x)^2=16`

`=> (3+x)^2=4^2`

`=>3+x=4`

`=>x=4-3`

`=>x=1`

Vậy: `x=1`

\(105+\left(3+x\right)^2=121\)

\(\Rightarrow\left(3+x\right)^2=121-105\)

\(\Rightarrow\left(3+x\right)^2=16\)

\(\Rightarrow\left(3+x\right)^2=\left(\pm4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}3+x=4\\3+x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4-3\\x=-4-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy \(x\in\left\{1;-7\right\}\)

Ư(120) thuộc {1;2;3;4;5;6;8;10;12;15;20;24;30;40;60;120}

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