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sorry nghe h tớ gửi quá 100 tin nhắn nên nó ko cho gửi
Bài 1
a)2711>818
b)6255>1257
c)536<1124
d)32n>23n
Bài 2
a)523<6.522
b)7.213>216
c)2115<275.498
a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
\(81\cdot\left(27+9^{15}\right):\left(3^5+3^{32}\right)\)
\(=\left(3^4\cdot3^3+3^4\cdot3^{30}\right):\left(3^5+3^{32}\right)\)
\(=\dfrac{3^7+3^{34}}{3^5+3^{32}}\)
\(=\dfrac{3^2\left(3^5+3^{32}\right)}{3^5+3^{32}}=9\)
\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)
\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\)
\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)
Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\)
\(-\left(-630\right)+915+\left(-630\right)+\left(-615\right)\)
\(=630+915-630-615\)
\(=915-615\)
\(=300\)
-(-630)+915+(-630)+(-615)
= 630 + 915 - 630 - 615
= (630 - 630) + (915 - 615)
= 0 + 300
= 300
Ta có: 915 = (32)15 = 330
2711 = (33)11 = 333
Vì 30 < 33 nên 330 < 333
Vậy 915 < 2711
9^15 = (3^2)^15 = 3^30
27^11= (3^3)^11 = 3^33
Vì 3^ 30< 3^33 nen 9^15< 28^11