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a) Ta có: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)
b) Ta có : \(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
\(\Rightarrow8192^7>3125^7\)\(\Rightarrow2^{91}>3^{35}\)
c) Ta có: \(99^{20}=99^{2.10}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)
a) \(2^{91}\)và \(5^{35}\)
Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192^7>3125^7\)nên \(2^{91}>5^{35}\)
b) \(3^{4000}\)và \(9^{2000}\)
Ta có :
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
Vì \(81^{1000}=81^{1000}\)nên \(3^{4000}=9^{2000}\)
\(2^{91}\)và \(5^{35}\)
Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192>3125\)nên \(2^{91}>5^{35}\)
\(3^{4000}\)và \(9^{2000}\)
Ta có :
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
Vì \(81=81\)nên \(3^{4000}=9^{2000}\)
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Mà 8^75 < 9^75 => 2^225<3^150
b) Ta có
2^91=(2^13)^7=8192^7
3^35=(3^5)^7=243^7
mà 8192^7<243^7=> 2^91<3^35
c) 3^4000=(3^2)^2000=9^2000
d) 2^332 < 2^333=2^3^111=8^111
3^223>3^222=9^111
=>2^332<3^223
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a) ta có :\(2^{24}=\left(2^2\right)^{12}=4^{12}\)
\(3^{36}=\left(3^2\right)^{12}=9^{12}\)
Vì \(4^{12}< 9^{12}\left(4< 9\right)\)
Nên bạn tự kết luận
b) ta có : \(10^{20}=\left(10^2\right)^{10}=100^{10}\)
Vì \(100^{10}>90^{10}\left(100>90\right)\)
Nên bạn tự kết luận
c) ta có : \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\left(8< 9\right)\)
Nên bạn tự kết luận
224=(22)12=412
336=(33)12=2712
Tự so sánh nhé
phần sau tương tự
a.\(2^{225}=\left(2^3\right)^{75}=8^{75}\left(1\right)\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow2^{225}< 3^{150}\)
b.\(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\left(1\right)\)
\(5^{35}=5^{7.5}=\left(5^5\right)^7=3125^7\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow5^{35}< 2^{91}\)
c.\(9999^{10}=\left(99.101\right)^{10}=99^{10}.101^{10}>99^{10}.99^{10}=99^{20}\)
\(\Rightarrow9999^{10}>99^{20}\)
Bài 2:
\(b.\)\(75^{20}=\left(3.5^2\right)^{20}=\left(3^{20}.5^{10}\right).5^{30}=\left[̣\left(3^2\right)^{10}.5^{10}\right].5^{30}=45^{10}.5^{30}\)
\(\Rightarrowđpcm\)
Bài 3:
a,\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=10^8\)
b. \(\frac{27^2}{25^3}=\frac{\left(3^3\right)^2}{\left(5^2\right)^3}=\frac{3^6}{5^6}=\left(\frac{3}{5}\right)^3\)\(:v\)
\(27^2.25^3=3^6.5^6=15^6\)( đề phòng you viết sai đề )
Ta có : \(2^{225}=\left[2^3\right]^{75}=8^{75}\)
\(3^{150}=\left[3^2\right]^{75}=9^{75}\)
Vì 8 < 9 => \(8^{75}< 9^{75}\)nên \(2^{225}< 3^{150}\)
Ta có : \(2^{91}>2^{90}=\left[2^5\right]^{18}=32^{18}>25^{18}=\left[5^2\right]^{18}=5^{36}>5^{35}\)
=> \(2^{91}>5^{35}\)
Ta có : \(9999^{10}=\left[99\cdot101\right]^{10}=99^{10}\cdot101^{10}>99^{10}\cdot99^{10}=99^{20}\)
Do đó : \(99^{20}< 9999^{10}\)
Cách khác : \(9999^{10}>9900^{10}=\left[99\cdot100\right]^{10}>\left[99^2\right]^{100}=99^{20}\)
Vậy : \(99^{20}< 9999^{10}\).
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Vì \(8^{75}< 9^{75}\Leftrightarrow2^{225}< 3^{150}\)
b)Ta có :
\(2^{91}=\left(2^{13}\right)^7=8193^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8193^7>3125^7\Leftrightarrow2^{91}>5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801^{10}>9999^{10}\Leftrightarrow99^{20}>9999^{10}\)
a) Ta có :
2225=(23)75=8752225=(23)75=875
3150=(32)75=9753150=(32)75=975
Vì 875<975⇔2225<3150875<975⇔2225<3150
b)Ta có :
291=(213)7=81937291=(213)7=81937
535=(55)7=31257535=(55)7=31257
Vì 81937>31257⇔291>53581937>31257⇔291>535
c) 9920=(992)10=9801109920=(992)10=980110
Vì 980110>999910⇔9920>999910980110>999910⇔9920>999910
a, Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
b, Ta có : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
\(\Rightarrow3^{2009}< 9^{1005}\)
c, Ta có : \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
a) Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì 9>8 nên 9111>8111
Vậy 3222>2333
b) Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Vì 2010>2009 nên 32010>32009
Vậy 91005>32009
c)Ta có:\(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 nên (99.99)10<(99.101)10
Vậy 9920<999910
a) \(2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8< 9\)\(\Rightarrow8^{111}< 9^{111}\)\(\Rightarrow2^{333}< 3^{222}\)
b) \(9^{1005}=\left(3^2\right)^{1005}=3^{2.1005}=3^{2010}>3^{2009}\)