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Ta có
3344=(3.11)44=344.1144=(34)11.1144=8111.1144
4433=(4.11)33=433.1133=(43)11.1133=6411.1133
=> 3344>4433
KL:
b) 52222=(52)1111=251111
25555=(25)1111=321111
=> 52222<25555
KL
Ta có :
- 9999=101.99\(\Rightarrow\)999910=(101.99)10=10110.9910
- 9920=9910+10=9910.9910
Vì 10110>9910\(\Leftrightarrow\)10110.9910>9910.9910\(\Leftrightarrow\)999910>9920
Vậy 999910>9920
227 = (23)9 = 89
318 = ( 32)9 = 99
Vì 9 > 8 nên : 99 > 89
Vậy suy ra: 318 > 227
\(8^{15}=\left(2^3\right)^{15}=2^{3.15}=2^{45}\\ 16^4=\left(2^4\right)^4=2^{4.4}=2^{16}\\ 2^{45}>2^{16}\Rightarrow8^{15}>16^4\)
\(a=\left[\left(-\dfrac{1}{2}\right)^5\right]^{107}=\left(-\dfrac{1}{32}\right)^{107}\)
\(b=\left[\left(-\dfrac{1}{3}\right)^3\right]^{107}=\left(-\dfrac{1}{27}\right)^{107}\)
mà -1/32>-1/27
nên a>b
a: \(8+\dfrac{5}{13}\simeq8,\left(384615\right)< 8,415...\)
b: \(-\dfrac{4}{7}=-0.\left(571428\right)\)
a) Ta có: \(\dfrac{34}{35}< 1;\dfrac{21}{20}>1\Rightarrow\dfrac{34}{35}< 1< \dfrac{21}{20}\)
Vậy \(\dfrac{34}{35}< \dfrac{21}{20}\)
a: \(\dfrac{34}{35}< 1< \dfrac{21}{20}\)
b: \(-\dfrac{123}{124}>-1>\dfrac{-321}{312}\)
c: \(\dfrac{1}{31}>\dfrac{1}{41}\)
\(\Leftrightarrow\dfrac{10}{31}+1>\dfrac{10}{41}+1\)
\(\Leftrightarrow\dfrac{41}{31}>\dfrac{51}{41}\)
a,\(2^{31}=2^{30}.2=\left(2^3\right)^{10}.2=8^{10}.2< 9^{10}.3=\left(3^2\right)^{10}.3=3^{20}.3=3^{21}\)
b,\(2^{99}=\left(2^3\right)^{33}=8^{33}>3^{21}\)
c,\(31^{14}< 32^{14}=\left(2^5\right)^{14}=2^{70}< 2^{72}=\left(2^4\right)^{18}=16^{18}< 17^{18}\)
d,\(63^{10}< 64^{10}=\left(2^6\right)^{10}=2^{60}< 2^{65}=\left(2^5\right)^{13}=32^{13}< 33^{13}\)