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Có: \(^{9^{36}}\)=\(^{\left(3^2\right)^{36}}\)=\(3^{72}\)
\(27^{10}\)=\(\left(3^3\right)^{10}\)=\(3^{30}\)
Vì 3 mũ 72 > 3 mũ 30 suy ra 9 mũ 36 > 27 mũ 10
ta có 1020=(102)10=10010>910
Vậy 1020>910
Chúc học tốt!
a,312 và 58
Ta có:312=(33)4=274
58=(52)4=254
Vì 274>254 nên 312>58
b,(0,6)9 và (0,9)6
Ta có:(0,9)6>(0,6)6 mà (0,6)6>(0,6)9
\(\Rightarrow\)(0,6)9<(0,9)6
c,52000 và 101000
Ta có:52000=(52)1000=251000>101000
\(\Rightarrow\)52000>101000
d,?????
\(\left(\dfrac{1}{27}\right)^{10}=\dfrac{1}{27^{10}}=\dfrac{1}{\left(3^3\right)^{10}}=\dfrac{1}{3^{30}}\)
\(\left(\dfrac{1}{81}\right)^7=\dfrac{1}{81^7}=\dfrac{1}{\left(3^4\right)^7}=\dfrac{1}{3^{28}}\)
Do \(3^{30}>3^{28}\Leftrightarrow\dfrac{1}{3^{30}}< \dfrac{1}{3^{28}}\)
\(\Leftrightarrow\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)
Ta có:
\(\left(\dfrac{1}{27}\right)^{10}=\left(\dfrac{1}{3^3}\right)^{10}=\left(\dfrac{1}{3}\right)^{30}\)
\(\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^5}\right)^7=\left(\dfrac{1}{3}\right)^{35}\)
Vì \(\left(\dfrac{1}{3}\right)^{35}>\left(\dfrac{1}{3}\right)^{30}\)
⇒\(\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)
Ta có :
527 = ( 53 )9 = 1259 < 1289 = ( 27 )9 = 263
=> 527 < 263
Mà 263 < 264 = ( 216 )4 = 655364 < 528 = ( 57 )4 = 781254
=> 527 < 263 < 528
a) Ta có :
\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)
\(\Rightarrow27^{27}>243^{13}\)
\(\Rightarrow-27^{27}< -243^{13}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)
b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)
c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)
\(\Rightarrow4^{50}>8^{30}\)
d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)
\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)
a) Ta có :
2727>2726=(272)13=72913>243132727>2726=(272)13=72913>24313
⇒2727>24313⇒2727>24313
⇒−2727<−24313⇒−2727<−24313
⇒(−27)27<(−243)13⇒(−27)27<(−243)13
b) (18)25>(18)26=(182)13=(164)13>(1128)13(81)25>(81)26=(821)13=(641)13>(1281)13
⇒(18)25>(1128)13⇒(81)25>(1281)13
⇒(−18)25<(−1128)13⇒(−81)25<(−1281)13
c) 450=(45)10=102410450=(45)10=102410
830=(83)10=51210<102410830=(83)10=51210<102410
⇒450>830⇒450>830
d) (19)17<(19)12<(127)12(91)17<(91)12<(271)12
⇒(19)17<(127)12⇒(91)17<(271)12
536=52*18=(52)18=2518
290=25*18=(25)18=3218
Vì 25<32 nên 2518<3218 hay 536<290
Vậy 536<290
9^36=(3^3)^36=3^108
27^10=(3^9)^10=3^90
Suy ra 9^36>27^10