\(A=-a+b-5a-5b\)

\(A=-\left(a-b\right)-5\left(a+b\right)\)

Thay a = -56 và b = 57 vào A ta có:

\(A=-\left(-56-57\right)-5\cdot\left(-56+57\right)\)

\(A=-\left(-113\right)-5\cdot1\)

\(A=113-5\)

\(A=108\)

a, 8¹⁰²

= (8⁴)²⁵.8²

= \(\overline{...6}\)²⁵.4

= \(\overline{..4}\)

b, 2017¹⁹⁹¹

= (2017⁴)⁴⁹⁷.2017³

= \(\overline{..1}\)⁴⁹⁷.\(\overline{..9}\)

= \(\overline{...9}\)

Ta có:

\(5566=2\cdot11^2\cdot23\)

\(1815=3\cdot11^2\cdot5\)

\(\Rightarrow BCNN\left(5566,1815\right)=2\cdot5\cdot3\cdot11^2\cdot23=83490\)

\(\Rightarrow BC\left(5566,1815\right)=B\left(83490\right)=\left\{0;83490;166980;250470;...\right\}\)

A = - 5²² - { - 222 - [ - 122 - (100 - 5²²) + 2022] }

A = - 5²² - { -222 - [- 122 - 100 + 5²² ] + 2022}

A = - 5²² - { -222 - { - 222 + 5²² } + 2022}

A = - 5²² - {- 222 + 222 - 5²² + 2022}

A = -5²² + 5²² - 2022

A = - 2022

B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)

B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2

B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2

B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)

B = \(\dfrac{2+3+4+...+21}{2}\)

B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)

B = \(\dfrac{23\times20:2}{2}\)

B = \(\dfrac{23\times10}{2}\)

B = 23 

(2n + 28) ⋮ (n + 3)    (n \(\ne\) -3)

(2n + 6 + 22) ⋮ (n + 3)

[2.(n + 3) + 22]⋮ (n + 3)

                 22  ⋮ (n + 30

(n + 3) \(\in\)  Ư(22) 

22 = 2.11 ⇒ Ư(22) = {-22; -11; -2; -1; 1; 2; 11; 22}

Lập bảng ta có:

Theo bảng trên ta có: 

n \(\in\) {-25; -14; -5; -4; -2; -1; 8; 19}

là sao bạn , ảo thế

ta có :

9¹⁸=9^3.6=(9³)⁶=27⁶

vì 12>6

=> 27¹²>27⁶

=>27¹²>9¹⁸

a/

2020.2021=(2019+1)(2022-1)=

=2019.2022-2019+2022-1=2019.2022+2>2019.2022

b/

\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)

c/

\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)

\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)

\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)

d/

\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)

\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)

\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)