b:\(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)

c: \(1+3+5+...+37\)

Số số hạng là:

(37-1):2+1=19(số)

Tổng là: \(\dfrac{38\cdot19}{2}=19\cdot19=361\)

\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{-2}{9}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{-2-5\cdot\left(-6\right)}{9}=\dfrac{-2+30}{9}=\dfrac{28}{9}\)

a: \(3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^2}\)

\(=\frac{3^2}{3^5}\cdot\frac{3^8}{3^2}=\frac{1}{3^3}\cdot3^6=3^3=27\)

b: \(4^6\cdot256^2\cdot2^4=\left(2^2\right)^6\cdot\left(2^8\right)^2\cdot2^4\)

\(=2^{12}\cdot2^{16}\cdot2^4=2^{12}\cdot2^{20}=2^{32}\)

c: \(4^6\cdot9^5+6^9\cdot120\)

\(=2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5\)

\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5\)

\(=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)

Ta có \(8^4\cdot3^{12}-6^{11}\)

\(=2^{12}\cdot3^{12}-2^{11}\cdot3^{11}\)

\(=2^{11}\cdot3^{11}\left(2\cdot3-1\right)=2^{11}\cdot3^{11}\cdot5\)

Ta có: \(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac45\)

d: \(4^2\cdot25^2+32\cdot125\)

\(=2^4\cdot5^4+2^5\cdot5^3\)

\(=2^4\cdot5^3\left(5+2\right)=2^4\cdot5^3\cdot7\)

\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}=2\cdot5\cdot7=70\)

a) (3⁵ . 3⁷) : 3¹⁰ + 5 . 2⁴ - 7³ : 7

= 3¹² : 3¹⁰ + 5.16 - 7²

= 3² + 80 - 49

= 9 + 31

= 40

b) (7⁵ + 7⁹) . (5⁴ + 5⁶) . (3³.3 - 9²)

= (7⁵ + 7⁹) . (5⁴ + 5⁶) . (81 - 81)

= (7⁵ + 7⁹) . (5⁴ + 5⁶) . 0

= 0

1.

a = -38

b = 154

c = -5871

d = -1600

a, -15 + (-22) - (-7) - |-8|

= -15 - 22 + 7 - 8

= -37 + 7 - 8

= -30 - 8

= - 38

a) (46+34)+(-80)=80+(-80)=0

b)2017\(^0\)=0

=> 5.(-9).2.(-7).(2017\(^0\))=0

c)115+85+35-46-35

=(35-35)+(115+85-46)

=154

d) 57.33-57.73

=57.(33-73)

=57.(-40)=-2280

`a, = 53 xx (12+172 -84)`

`= 53 xx 100`

`=5300`

`b,= 2018 xx (91-45-46)`

`= 2018 xx 0`

`=0`

15,8/3,5=158/35=4 dư 18

35:17=2 dư 1

26,95:2,1=77/6=12 dư 5

46:15=3 dư 1

đặt tính ra

a: \(\left(-2\right)^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot1^{2019}\)

\(=-8-45:9+1\)

\(=-8-5+1\)

=-13+1

=-12

b: \(11^{25}:11^{13}-3^5:\left(1^{10}+2^3\right)-60\)

\(=11^{25-13}-3^5:3^2-60\)

\(=11^{12}-27-60\)

\(=11^{12}-87\)

EM CẢM ƠN !