Ta có :

            S = 5 + 5² +5³ +5⁴ +.... + 5¹⁰⁰        có (100 - 1) : 1 + 1 = 100 số hạng

           S = (5 + 5²) + (5³ + 5⁴) + ....... + (5⁹⁹ + 5¹⁰⁰)

          S = 5 . (1 + 5) + 5³ . (1 + 5) + .... + 5⁹⁹ . (1 + 5)

          S = 5 . 6 + 5³ . 6 + ..... + 5⁹⁹ . 6

         S = 6 . (5 + 5³ + ..... + 5⁹⁹)

        Vì 6 chia hết cho 6 nên S chia hết cho 6 (ĐPCM)

        Ủng hộ mk nha !! ^_^

\(S=5+5^2+5^3+...+5^{100}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+.....+\left(5^{99}+5^{100}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+......+5^{99}\left(1+5\right)\)

\(=\left(1+5\right)\left(5+5^3+.....+5^{99}\right)\)

\(=6\left(5+5^3+....+5^{99}\right)\)

1: 

Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)

2) Để A là số nguyên dương thì 

\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{6;12\right\}\)

Giải:

1)D=3/5.7+3/7.9+3/9.11+...+3/53.55

   D=3/2.(2/5.7+2/7.9+2/9.11+...+2/53.55)

   D=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55)

   D=3/2.(1/5-1/55)

   D=3/2.2/11

   D=3/11

S=(1+2)+(2^2+2^3)+(2^4+2^5)+....+(2^99+2^100)

S=3+3.2^2+3.2^4+.....+3.2^99

S=3.(2^2+2^4+.....+2^99)

Vì 3 chia hết 3=>3.(2^2+2^4+....+2^99)

=>S chia hết 3

2S=2+2^2+2^3+2^4+.....+2^101

2S-S=(2+2^2+2^3+2^4+....+2^101)-(1+2+2^2+2^3+2^4+....+2^100)

S=2^101-1

S+1=2^101-1+1=2^101

=>x=101

tích đúng cho mình nha

S = (2¹+2²)+(2³+2⁴)+...+(2⁹⁹+2¹⁰⁰)

S = 2.(1+2)+2³.(1+2)+...+2⁹⁹.(1+2)

S = 2.3+2³.3+...+2⁹⁹.3

S = 3.(2+2³+...+2⁹⁹)

=> S chia hết cho 3

S = (2¹+2²+2³+2⁴)+(2⁵+2⁶+2⁷+2⁸)+...+(2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

S = 2.(1+2+4+16)+2⁵.(1+2+4+16)+...+2⁹⁷.(1+2+4+16)

S = 2.15+2⁵.15+...+2⁹⁷.15

S = 15.(2+2⁵+...+2⁹⁷)

=> S chia hết cho 15

Bài dễ ợt ai mà chẳng làm được

ukm

thanks

Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99 
           =2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
           =2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
           =1+0+0+0+...+0+1/99
           =1+1/99
           =100/99
Mà 100/99>1.Suy ra S>1
   Vậy S>1
            

S=1-1/3 + 1/3 - 1/5 + ... + 1/97 - 1/99

=1 - 1/99 => S<1