*) Chứng minh A ⋮ 5

Ta có:

A = 4¹⁹ + 4¹⁸ + ... + 4² + 4 + 1

= (4¹⁹ + 4¹⁸) + ... + (4³ + 4²) + (4 + 1)

= 4¹⁸.(4 + 1) + ... + 4².(4 + 1) + (4 + 1)

= 4¹⁸.5 + ... + 4².5 + 5

= 5(4¹⁸ + ... + 4² + 1) ⋮ 5

Vậy A ⋮ 5

*) Chứng minh A ⋮ 17

Ta có:

4¹⁹ + 4¹⁸ + ... + 4² + 4 + 1

= 4¹⁹ + 4¹⁸ + 4¹⁷ + 4¹⁶ + ... + 4³ + 4² + 4 + 1

= (4¹⁹ + 4¹⁸ + 4¹⁷ + 4¹⁶) + ... + (4³ + 4² + 4 + 1)

= 4¹⁶(4³ + 4² + 4 + 1) + ... + (4³ + 4² + 4 + 1)

= 4¹⁶.85 + ... + 85

= 85.(4¹⁶ + ... + 1) ⋮ 17 (vì 85 ⋮ 17)

Vậy A ⋮ 17

sorry bn tui hỏng bt

ko biết đừng nhắn bn ơi

\(4S=4+4^2+4^3+4^4+...+4^{100}\)

\(3S=4S-S=4^{100}-1\Rightarrow3S+1=4^{100}\)

Ta có \(32^{20}=\left(2^5\right)^{20}=2^{100}\)

\(\Rightarrow4^{100}>2^{100}\Rightarrow3S+1>32^{20}\)

a)419  x 2 + 419 x 3 + 419 x 5

= 419 x ( 2 + 3 + 5 )

= 419 x 10

= 4190

b) 7020 x 7 - 7020 x 3 - 7020 x 4

= 7020 x ( 7 - 3 - 4 )

 = 7020 x 1

= 7020

a, 419 x ( 2 + 3 +5)

= 419 x 10 = 4190.

b, 7020 x ( 7 - 3 - 4)

= 7020 x 0 = 0.

a)419*2+419*3+419*5      

=419*(2+3+5)

=419*10

=4190

b)7020*7-7020*3-7020*4

=7020*(7-3-4)

=7020*0

=0

k cho mk nha bn

THANK YOU VERY MUCH

1201 > 999        24 601 > 2461

43 658 <43 690        138 579 < 138 701

5178 = 5100 + 78       520 000 > 419 999

a) (689 + 875) + 125

= 689 + (875 + 125)

= 689 + 1000 = 1689.

581 + (878 + 419)

= (581 + 419) = 878

= 1000 + 878 = 1878.

b)

c) 5,87 + 28,69 + 4,13

= (5,87 + 4,13) + 28,69

= 10 + 28,69

= 38,69.

83,75 + 46,98 + 6,25

= (83,75 + 6,25) + 46,98

= 90 + 46,98

= 136,98.

a)(689+875)+125=689+(875+125)=689+1000=1689

581 + (878+419)=581+419+878=1000+878=1878

b)(2/7+4/9)+5/7=2/7+5/7+4/9=7/7+4/9=1+4/9=9/9+4/9=13/9

17/11+(7/15+5/11)=17/11+5/11+7/15=2+7/15=30/15+7/15=3/15

c)5,87+28,69+4,13=5,87+4,13+28,69=10+28,69=38,69

83,75+46,98+6,25=83,75+6,25+46,98=90+46,98=136,98

\(4^0+4^1+4^2+4^3+...+4^{35}\\ 4S=4^1+4^2+4^3+4^4+...+4^{36}\\ 4S-S=\left(4^1+4^2+4^3+4^4+...+4^{36}\right)-\left(4^0+4^1+4^2+4^3+...+4^{35}\right)\\ 3S=4^{36}-1=64^{12}-1\\ Vì64^{12}-1< 64^{12}\\ \Rightarrow3S< 64^{12}\)

Ta có: \(64^{12}=\left(4^3\right)^{12}=4^{36}\)

\(S=4^0+4^1+...+4^{34}+4^{35}\)

\(\Rightarrow4S=4^1+4^2+...+4^{35}+4^{36}\)

\(\Rightarrow4S-S=4^{36}-4^0\)

\(\Rightarrow3S=4^{36}-1< 4^{36}\)

Vậy \(3S< 64^{12}\)

a, ta có :

 \(\hept{\begin{cases}\frac{6}{7}< 1\\\frac{11}{10}>1\end{cases}}\Rightarrow\frac{6}{7}< \frac{11}{10}\)

b, ta có :

\(\hept{\begin{cases}\frac{-5}{7}< 0\\\frac{2}{7}>0\end{cases}}\Rightarrow\frac{-5}{7}>\frac{2}{7}\)

c, ta có :

\(\hept{\begin{cases}\frac{419}{-723}< 0\\\frac{-697}{-313}>0\end{cases}}\Rightarrow\frac{419}{-723}< \frac{-679}{-313}\)

a]6/7 va 11/10 

60/70 và 77/70 vay 11/10 lớn hơn 6/7 

a) \(\frac{6}{7}\) và \(\frac{11}{10}\)

Vì \(\frac{6}{7}< 1;1< \frac{11}{10}\)

\(\Rightarrow\frac{6}{7}< \frac{11}{10}\)

Vậy \(\frac{6}{7}< \frac{11}{10}\)

a,Ta có:

6/7< 1< 11/10

=> 6/7< 11/10

b,Ta có:

-5/17< 0< 2/7

=> -5/7< 2/7

419/-723 > o > -697/-313