\(2+2^2+2^3+2^4+...+2^9+2^{10}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=3\left(2+2^3+...+2^9\right)⋮3\)

Số số hạng của dãy số A là:

\(\dfrac{199-1}{2}+1=\dfrac{198}{2}+1=100\left(số\right)\)

Tổng của dãy số A là:

\(\left(1+199\right)\cdot\dfrac{100}{2}=100^2=10000\)

Số số hạng của dãy số B là:

\(\dfrac{999-100}{1}+1=899+1=900\left(số\right)\)

Tổng của dãy số là: \(B=\left(999+100\right)\cdot\dfrac{900}{2}=494550\)

\(A+B=10000+494550=504550\)

cc

\(C=\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+...+\dfrac{1}{638}\)

\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{3190}\)

\(=\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+...+\dfrac{5}{55\cdot58}\)

\(=\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{55\cdot58}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{55}-\dfrac{1}{58}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{58}\right)=\dfrac{5}{3}\cdot\dfrac{27}{116}=\dfrac{5\cdot9}{116}=\dfrac{45}{116}\)

\(C=\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+...+\dfrac{1}{638}\\ =\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{3190}\\ =\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+\dfrac{5}{10\cdot13}+...+\dfrac{5}{55\cdot58}\\ =\dfrac{5}{3}\cdot\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{55\cdot58}\right)\\ =\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{55}+\dfrac{1}{55}-\dfrac{1}{58}\right)\\ =\dfrac{5}{3}\cdot\left(\dfrac{1}{4}-\dfrac{1}{58}\right)\\ =\dfrac{5}{3}\cdot\dfrac{27}{116}\\ =\dfrac{45}{116}\)

1: \(1935⋮5;540⋮5;270⋮5\)

Do đó: \(1935-540+270⋮5\)

\(1935⋮9;540⋮9;270⋮9\)

Do đó: \(1935-540+270⋮9\)

2: \(5^{3x-1}-5^{2x+1}=0\)

=>\(5^{3x-1}=5^{2x+1}\)

=>3x-1=2x+1

=>3x-2x=1+1

=>x=2

y chia 19 được thương là 20, dư là 8

=>\(y=19\cdot20+8=380+8=388\)

\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=5-1\\ \Rightarrow2x=4\\ \Rightarrow x=4:2\\ \Rightarrow x=2\)

\(\left(2x+1\right)^3=125\)

=>\(\left(2x+1\right)^3=5^3\)

=>2x+1=5

=>2x=5-1=4

=>\(x=\dfrac{4}{2}=2\)

Để `5n+22 \vdots n+3,` ta có:

`5n +22 \vdots n+3`

`=> 5n + 15 + 7 \vdots n + 3`

`=> 5 (n + 3)  + 7 \vdots n + 3`

Vì:: `5 ( n + 3)\vdots n + 3 -> n  + 3 in Ư(7)={+-1;+-7}`

`=> n = {-2;-4;4;-10}`

Vậy: `n = {-2;-4;4;-10}` thì `5n + 22 \vdots n+3`

\(5n+22⋮n+3\\ \Leftrightarrow5n+15+7⋮n+3\\ \Leftrightarrow7⋮n+3\text{ }\left(\text{Vì 5n + 14 ⋮ n + 3}\right)\\ \Leftrightarrow n+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}n+3=1\\n+3=-1\\n+3=7\\n+3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=-4\\n=4\\n=-10\end{matrix}\right.\)

Vậy \(n\in\left\{-2;-4;4;-10\right\}\)

D={0;4;8;12;16;20}

D={x\(⋮\)4; x<21}

Cách 1:

\(D=\left\{0;4;8;12;16;20\right\}\)

Cách 2:

\(D=\left\{x\in N|x⋮4,x< 21\right\}\)

\(\left(3\cdot4\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot4^{11}-16^9\right)\\ =\left(3\cdot2^2\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot2^{22}-2^{36}\right)\\ =3^2\cdot2^4\cdot2^{32}:\left(11\cdot2^{35}-2^{36}\right)\\ =3^2\cdot2^{36}:\left[2^{35}\cdot\left(11-2\right)\right]\\ =9\cdot2^{36}:\left(2^{35}\cdot9\right)\\ =9\cdot2^{36}:2^{35}:9\\ =2\)

Để `(x+3)\vdots(x+1),` ta có:

`(x+3)\vdots(x+1)`

`=> (x+1)+2\vdots(x+1)`

Vì: `(x+1)\vdots(x+1)` \(\rightarrow\) `(x+1)` thuộc `Ư(2) = {+-1;+-2}`

`=> x = {0;-2;1;-3}`

Vậy: `x={0;-2;1;-3}` thì `(x+3)\vdots(x+1)`

(x+3)⋮(x+1)

x+1+2⋮x+1

2⋮x+1 (Vì x+1⋮x+1)

=> x+1 thuộc Ư(2) = {-1; 1; 2; -2}

=> x thuộc {-2; 0; 1; -3}

Vậy x thuộc {-2, 0; 1; -3}