b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)

\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)

\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)

\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)

\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)

\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)

=2745427451

c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)

\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)

=0+9

=9

b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)

\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)

\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)

\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)

\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)

\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)

=2745427451

c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)

\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)

=0+9

=9

a) 13.7 + 7.7

= (13 + 7).7

= 20 . 7

= 140

b) 2³ . 2016 . 5²

= 8 . 2016 . 25

= 200 . 2016

= 403200

c) 3² . 15 - 5.9

= 9.15 - 5.9

= (15 - 5).9

= 10.9

= 90

d) 13 + 15 + ........... + 57 + 59

Áp dụng công thức tính dãy số ta có :

\(\frac{\left[\left(59-13\right):2+1\right].\left(59+13\right)}{2}=\frac{24.72}{2}=864\)

a) 13 . 7 + 7 . 7

= (13 + 7) . 7

= 20 . 7

= 140

b) 2³ . 2016 . 5²

= 8 . 2016 . 25

= ( 8 . 25 ) . 2016

= 200 . 2016

= 403200 

c) 3² . 15 - 5 . 9

= 9 . 15 - 45

= 9 . 15 - 15 . 3

= ( 9 - 3 ) . 15

= 6 . 15

= 90

d) 13 + 15 + ........ + 57 + 59

Dãy số trên có số số hạng là:

( 59 - 13 ) : 2 + 1= 24 ( số hạng )

Tổng của dãy số trên là:

( 59 + 13 ) x 24 : 2 = 864

MK lm bài 1:

\(\left(a-b\right)\left(a+b\right)=a^2+ab-ab-b^2=a^2-b^2\)

Bài 1:

\(\left(a-b\right)\left(a+b\right)=a^2+ab-ab-b^2\)

\(=a^2-b^2\)

a) 17 + 15 + (-27)

= (17 - 27) + 15

= -10 + 15

= 5

b) 13.75 + 25.13 - 300

= 13.(75 + 26) - 300

= 13.100 - 300

= 1300 - 300

= 1000

c) 80 - (4.5² - 5.2³) + 2021⁰

= 80 - (4.25 - 5.8) + 1

= 80 - (100 - 40) + 1

= 80 - 60 + 1

= 20 + 1

= 21

a)2.18.14+3.17.12+4.69.9
=(2.18).14+(12.3).17+(4.9).69
=36.14+36.17+36.69
=36.(14+17+69)
=36.100
=3600

b)1+(-3)+5+(-7)+9+(-11)+13+(-15)
=[1+(-3)]+[5+(-7)]+[9+(-11)]+[13+(-15)]
=(-2)+(-2)+(-2)+(-2)
=(-8)

\(\left[504-\left(5^2.8+70\right):3^3+6\right]:25\)

\(\left[504-\left(25.8+70\right):27+6\right]:25\)

\(\left[504-\left(200+70\right):27+6\right]:25\)

\(\left[504-270:27+6\right]:25\)

\(\left[504-10+6\right]:25\)

\(\left[494+6\right]:25\)

\(500:25\)

= 20

\(\left[504-(5^2\cdot8+70):3^3+6\right]:25\)

\(=\left[504-(25\cdot8+70):27+6\right]:25\)

\(=\left[504-(200+70):27+6\right]:25\)

\(=\left[504-270:27+6\right]:25\)

\(=\left[504-10+6\right]:25\)

\(=\left[494+6\right]:25\)

\(=500:25=20\)

\(\text{Ko chắc }:)\)

a. (-7).(-3).17.(-1)²

= (-7).(-3).17.1

= 21.17.1

=357.1

=357