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a) Ta có:
\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)
Mà: \(8< 9\)
\(\Rightarrow8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) Ta có:
\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)
Mà: \(243< 343\)
\(\Rightarrow243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
c) Ta có:
\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)
\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)
Mà: \(2< 3\)
\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)
\(\Rightarrow8^5< 3\cdot4^7\)
d) Ta có:
\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)
\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)
Mà: \(8242408>91809\)
\(\Rightarrow8242408^{101}>91809^{101}\)
\(\Rightarrow202^{303}>303^{202}\)
Câu 1.9920và 999910
=(992)10=980110
Vậy 980110<999910 suy ra 9920<999910
Câu 2. 3500và 7300
3500=(35)100=243100
7300=(73)100=343100
Vậy 243100<343100 => 3500<7300
a)\(27^2\)và \(4^6\)
\(27^2=\left(3^3\right)^2\)
\(4^6=\left(2^3\right)^2\)
\(3^3>2^3\)
b) \(3^{500}=\left(3^5\right)^{100}\)
\(7^{300}=\left(7^3\right)^{100}\)
\(7^3=343\)
\(3^5=243\)
\(\Rightarrow3^{500}< 7^{300}\)
c) \(8^5=4^5\cdot2^5\)
\(3\cdot4^7=3\cdot4^2\cdot4^5\)
\(3\cdot4^2>2^5\)
\(3\cdot4\cdot4=2\cdot2\cdot2\cdot2\cdot3>2\cdot2\cdot2\cdot2\cdot2\)
\(8^5< 3\cdot4^7\)
d) \(202^{303}=\left(202^3\right)^{101}\)
\(303^{202}=\left(303^2\right)^{101}\)
\(202^3>303^2\)
Nên
a) 514 = 52.7= (52)7= 257
321= 33.7= (33)7= 277
Mà 25 < 27
=> 514<321
b) 433= (43)11 = 6411
344= (34)11 = 8111
=> 433<344
c) 2161 > 2160 = (24)40= 1640
Mà 1340< 1640 < 2161
=> 1340<2161
d) 1715= 173.5= (173)5 = 49135
3110= (312)5= 9615
Mà 4913 > 961
=> 1715> 3110
e) 2255 = (25)45 = 3245
3180 = (34)45 = 8145
=> 2255 < 3180
g) 202303 = (2023)101 = (1013.23)101= (1013.8)101= (1012.8.101)101= (1012.808)101
303202 = (3032)101= (32.1012)101= (1012.9) 101
Vì 1012.9 < 1012.808
=> 202303>303202
`@` `\text {Ans}`
`\downarrow`
\(202^{303}\text{ và }303^{202}\)
Ta có:
\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}\)
\(303^{202}=303^{101\cdot2}=\left(303^2\right)^{101}\)
So sánh `202^3` và `303^2`, ta có:
`202^3 = (2*101)^3 = 2^3 * 101^3 = 8 * 101^3 = 8* 101^2 * 101 = 808*101^2`
`303^2 = (3*101)^2 = 3^2 * 101^2 = 9 * 101^2`
Vì `9 < 808 \Rightarrow 9*101^2 < 808*101^2`
`\Rightarrow`\(202^{303}>303^{202}\)
Vậy, \(202^{303}>303^{202}.\)
a, 2^300 < 3^200
b, 3^500 < 7^300
c, 8^5 > 3.4^7
d, 202^303 < 303^202
a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)
\(\Rightarrow\text{ }4^{100}< 2^{202}\)
b, \(3^0=1< 5^8\)
\(3^0< 5^8\)
c, \(\left(0,6\right)^0=1\)
\(\left(-0,9\right)^6=\left(0,9\right)^6\)
\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)
d,
e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)
\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)
Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)
a/
\(2^{1050}=\left(2^2\right)^{525}=4^{525}< 5^{525}< 5^{540}\)
b/
\(2^{161}>2^{160}=\left(2^4\right)^{40}=16^{40}>13^{40}\)
c/
\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}>2^{55}=\left(2^5\right)^{11}=32^{11}>31^{11}\)