Ta có A = 2019.2021.a = (2020 – 1)(2020 + 1)a = ( 2020 2 – 1)a

Và B   =   ( 2019 2   +   2 . 2019   +   1 ) a   =   ( 2019   +   1 ) 2 a   =   2020 2 a

Vì 2020 2   –   1   <   2020 2 và a > 0 nên ( 2020 2   –   1 ) a   <   2020 2 a hay A < B

Đáp án cần chọn là: D

= 3029876 nha

Hok tot

Kb hh

20192 + 982 + 987 = 22 161

 Chúc hok tốt

Đáp án :

22161

hok tốt

_________________

a) \(153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)

b)

\(\left(2020^2-2019^2\right)+\left(2018^2-2017^2\right)+...+\left(2^2-1^2\right)\\ =\left(2020+2019\right)\left(2020-2019\right)+\left(2018+2017\right)\left(2018-2017\right)+...+\left(2+1\right)\left(2-1\right)\\ =2020+2019+2018+2017+...+2+1\\ =\dfrac{\left(2020+1\right)2020}{2}=2041210\)

Lời giải:

a. $153^2-53^2=(153-53)(153+53)=100.206=20600$

b. 

$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$

$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$

$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$

$=2020+2019+2018+2017+...+2+1$

$=\frac{2020.2021}{2}=2041210$

a) 153²-53²=(153-53)(153+53)=100.206=20600

\(A=2019^2-\left(2019-3\right)\left(2019+3\right)\\ A=2019-2019^2-2019\cdot3+2019\cdot3+3^2\\ A=9\)

1. 

$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$

2.

$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$

3. Không phù hợp để tính nhanh 

4. 

$=15^8-(15^8-1)=1$

5.

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$

$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$

$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$

6:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)

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