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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 \(2^{225}=8^{75}< 9^{75}=3^{150}\)
b: \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=5^{35}\)
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 )
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, 2225 = 215.15= ( 215)15 = 3276815
3150 = 310.15 = ( 310)15 = 5904915
Dễ thấy 32768 < 59049 nên 2225 < 3150
9920=(992)10=980110
vì 9801<9999
nên 980110<999910 hay
9920<999910
a) Ta có 9920=(992)10=980110
Vì 980110<999910 nên =>9920<999910
Ta có 231= 2.230=2.(23)10=2.810
321= 3.320=3.(32)10=3.910
Vì 2<3 ;810< 910 nên =>2.810<3.910
hay 231<321
d) ta có 17^20=(17^4)5=83521^5
vì 71^5<83521^5 nên => 71^5<17^20
- Câu c mình ra rùi nhưng số to lăm nên phải nghĩ cách khác đã
a, 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975 nên 2225 < 3150
b, 334 > 330 = (33)10 = 2710
521 > 520 = (52)10 = 2510
Vì 2710 > 2510 => 330 > 520 => 334 > 521
c, 321 > 320 = (32)10 = 910
231 > 230 = (23)10 = 810
Vì 910 > 810 => 321 > 231
d, 291 > 290 = (25)18 = 3218
535 < 536 = (52)18 = 2518
Vì 3218 > 2518 => 291 > 535
e, 9920 = (992)10 = 980110 < 999910
f, 128.912 = 38.48.324 = 332.212
1816 = 216.916 = 216.332
Vì 332 . 212 < 216.332 => 128.912 < 1816
g, 7520 = 2520.320 = 540.320
4510.530 = 510.910.530 = 540.320
Vậy 7520 = 4510.530
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: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)
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 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)