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5\(^{300}\)=25\(^{150}\)
3\(^{453}\)=27\(^{151}\)=27.27\(^{150}\)
vì 25\(^{150}\)<27.27\(^{150}\)
\(\Rightarrow\)5\(^{300}\)<3\(^{453}\)
31\(^{11}\)<32\(^{11}\)=(2\(^5\))\(^{11}\)=2\(^{55}\)
31\(^{11}\)<2\(^{55}\)
17\(^{14}\)>16\(^{14}\)=2\(^{56}\)
31\(^{11}\)<2\(^{55}\)<2\(^{56}\)<17\(^{14}\)
\(\Rightarrow\)31\(^{11}\)<17\(^{14}\)
333\(^{444}\)=3\(^{444}\).111\(^{444}\)
444\(^{333}\)=4\(^{333}\).111\(^{333}\)
ta có 3\(^{444}\)=81\(^{111}\)
4\(^{333}\)=64\(^{111}\)
\(\Rightarrow\)3\(^{444}\)>4\(^{333}\)(81\(^{111}\)>64\(^{111}\))
111\(^{444}\)>111\(^{333}\)
3\(^{444}\).111\(^{444}\)>4\(^{333}\).111\(^{333}\)
Vậy 333\(^{444}\)>444\(^{333}\)
nảy mình làm thiếu 1 câu bây giờ bù nhá
Bài 1 so sánh
333444 và 444333
Bài 2 so sánh
a) 321 và 231
b) 1990 10 + 19909 và 199110
c) 10750 và 7375
333444 và 444333
ta có : 333444 = ( 3334 )111 =12296370321111
444333 = ( 4443 )111 = 87528384111
vì 12296370321 > 87528384
=> 333444 > 444333
a) Ta có : \(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
mà \(1000< 1024\)
\(\Rightarrow1000^{10}< 1024^{10}\)
\(\Rightarrow10^{30}< 2^{100}\)
b) Ta có : \(333^{444}=\left(111.3\right)^{444}=111^{444}.3^{444}=111^{444}.\left(3^4\right)^{111}=111^{444}.81^{111}\)
\(444^{333}=\left(111.4\right)^{333}=111^{333}.4^{333}=111^{333}.\left(4^3\right)^{111}=111^{333}.64^{111}\)
mà \(444>333\Rightarrow111^{444}>111^{333}\)
và \(81>64\Rightarrow81^{111}>64^{111}\)
\(\Rightarrow111^{444}.81^{111}>111^{333}.64^{111}\)
\(\Rightarrow333^{444}>444^{333}\)
c) Ta có : \(2^{161}>2^{160}=\left(2^4\right)^{40}=16^{40}>13^{40}\)
\(\Rightarrow2^{161}>13^{40}\)
d) Ta có : \(3^{453}>3^{450}=\left(3^3\right)^{150}=27^{150}>25^{150}=\left(5^2\right)^{150}=5^{300}\)
\(\Rightarrow3^{453}>5^{300}\)
Bài2: a. 3500= (35).100=243100
7300= (73).100= 147100. Mà 243> 147 => 243100> 147100. Vây 3500> 7300
b.
2a.
3^500=(3^5)^100=243^100
7^300=(7^3)^100=343^100
Ta thấy :243^100<343^100 suy ra:3^500<7^300
a/ \(2A=2+2^2+2^3+2^4+...+2^{2011}\)
\(A=2A-A=2^{2011}-2^0=2^{2011}-1=B\)
b/ \(A=2009.2011=\left(2010-1\right)\left(2010+1\right)=2010^2-1< B=2010^2\)
c/
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow11^{24}=121^{12}< 125^{12}=5^{36}\)
d/
\(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}>5^{20}=625^5\)
e/
\(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n< 9^n=3^{2n}\)
f/
\(6.5^{22}>5.5^{22}=5^{23}\)
g/
\(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)
\(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)
\(\Rightarrow333^{444}>444^{333}\)
e, Ta có:
\(9^{20}=3^{40}\)
\(27^{13}=3^{39}\)
Vì \(\hept{\begin{cases}3=3\\40>39\end{cases}\Rightarrow}3^{40}>3^{39}\)
Vậy 920>2713