(2020+27) - [ 2020 + (-73) ] - 129
= 2047 - 1947 - 129
= 100 - 129 = -29

cảm ơn ạ

À câu này giải theo cách này e nhé!

E tham khảo nha:(-- được xem là +)

=>(2020+27)+(2020-73)-129

=2047+1947-129=3865

BÀi 1:

a: (-243+345)-(257+(-55)-129)

=-243+345-257+55+129

=(-243-257)+(345+55)+129

=-500+400+129

=-100+129

=29

b: \(\left(-2\right)^3-45:\left(-3\right)^2-\left(-2019\right)^0\cdot\left(-1\right)^{2019}\)

\(=-8-\frac{45}{9}-1\cdot\left(-1\right)\)

=-8-5+1

=-13+1

=-12

c: \(\left\lbrack27\cdot\left(-67\right)+33\cdot\left(-27\right)\right\rbrack:\left(-30\right)\)

\(=\frac{-27\left(67+33\right)}{-30}\)

\(=\frac{9}{10}\cdot100=90\)

d: \(\left(1^2+2^2+\cdots+100^2\right)\left(3^4-9^2\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)\left(81-81\right)\)

=0

e: \(53\cdot47+53^2-2021^0\)

\(=53\left(53+47\right)-1\)

=5300-1

=5299

i: 38+(-21)+(-58)+(-6)

=(38-58)+(-21-6)

=-20-27

=-47

h: (2020+27)-[2020+(-73)]-129

=2020+27-2020+73-129

=100-129

=-29

a: 21^15=3^15*7^15

27^5*49^8=3^15*7^14

mà 15>14

nên 21^15>27^5*49^8

b: \(2020^{2020}-2020^{2019}=2020^{2019}\left(2020-1\right)=2020^{2019}\cdot2019\)

\(2020^{2019}-2020^{2018}=2020^{2018}\cdot2019\)

mà 2019>2018

nên 2020^2020-2020^2019>2020^2019-2020^2018

cảm ơn ạ

\(a.=\dfrac{2019}{2020}\times\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\\ =\dfrac{2019}{2020}\times1=\dfrac{2019}{2020}\\ b.=\dfrac{25}{27}\times\left(\dfrac{17}{14}-\dfrac{1}{14}-\dfrac{2}{14}\right)\\ =\dfrac{25}{27}\times1=\dfrac{25}{27}\)

27 x 129 - 29 x 27 + 100 x 73

= 27 x ( 129 - 29 ) + 100 x 73

=27 x 100 + 100 x 73

=100 x (27 + 73 )

= 100 x 100

= 10000

k mình nha

I love you

\(27.\left(129-29\right)+100.73\)

\(27.100+100.73\)

\(100.\left(27+73\right)\)

\(100.100=10.000\)

tích ak