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\(9^8x5^{16}=\left(3^2\right)^8x5^{16}=3^{16}x5^{16}=\left(3x5\right)^{16}=15^{16}\)
Vì 15 < 19 => 1516 < 1916 < 1920
=> 1516 < 1920
Vậy 98x516 < 1920
9^8.5^16=81^4.(5^4)^4=175625^4
19^20=2476099^4
=>(9^8).(5^16)<19^20
1717/8585=17.101/85.101=17/85=1/5=13/65<13/51=13.101/51.101=1313/5151
Ta có 202^303= 202^ 3. 101= (202^3)^101=8242408^101
303^202=303^2.101 =(303^2)^101= 91809^101
vÌ 8242408 >91809 nên _______
vậy_____
A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)
10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2
B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)
10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2
10A > 2 > 10B ⇒ 10A>10B ⇒ A>B
8*9 lớn hơn 9*8
tk mình nha
8.9=9.8
easy