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Ta có:
\(\left(-5\right)^{30}=\left(-5^3\right)^{10}=\left(-125\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=\left(-3^5\right)^{10}=\left(-81\right)^{10}=81^{10}\)
Vì \(125^{10}>81^{10}\)
⇒\(\left(-5\right)^{30}>\left(-3\right)^{50}\)
\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)
Ta có: \(\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\)
\(\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\)
Vì \(\left(-125\right)^{10}< \left(-243\right)^{10}\) nên \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
1,1020và 9010
ta có:+,1020=(102)10=10010
+,9010=9010
vì 10010>9010=>1020>9010
2,(1/16)10 và (1/2)50
ta có:+, (1/16)10=(1/16)10
+,(1/2)50=(1/25)10=(1/32)10
vì (1/16)10>(1/32)10=>(1/16)10>(1/2)50
k mik nhé
\(a,\) \(10^{20}=10^{10+10}=10^{10}.10^{10}\)
\(90^{10}=9^{10}.10^{10}\)
Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)
\(\Rightarrow10^{20}>90^{10}\)
Vậy \(10^{20}>90^{10}\)
\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)
Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
~~~~~~~~~~Hok tốt~~~~~~~~~~~
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
\(\left(\dfrac{1}{2}\right)^{50}=\left[\left(\dfrac{1}{2}\right)^5\right]^{10}=\left(\dfrac{1}{32}\right)^{10}\)
1/12>1/32
=>(1/12)^10>(1/32)^10
=>(1/12)^10>(1/2)^50
Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)
\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}=\dfrac{1}{\left(2^5\right)^{10}}=\dfrac{1}{32^{10}}\)
Do \(12< 32\Rightarrow12^{10}< 32^{10}\)
\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)
\(2^{50}=\left(2^5\right)^{10}=32^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Suy ra: 250 > 520
b)
\(9^{200}=\left(9^2\right)^{100}=81^{100}\)
Suy ra: 99100 > 81100
\(3^{400^{100}}\)và \(4^{500^{50}}\)
\(\Rightarrow3^{\left(400^2\right)^{50}}\Leftrightarrow3^{160000^{50}}\)
\(\Rightarrow\left(3^{320}\right)^{500^{50}}\)
mà :\(3^{320}>4\)
\(\Rightarrow3^{400^{100}}>4^{500^{50}}\)
\(\left(-3\right)^{50}\)và \(\left(-50\right)^{30}\)
\(\Leftrightarrow\)\(\left(-3\right)^{50}=\left(-3^5\right)^{10}=-243^{10}\)
\(\Leftrightarrow\)\(\left(-50\right)^{30}=\left(-50^3\right)^{10}=-125000^{10}\)
Vì số mũ lẻ luôn âm và số mũ chắc luôn dương và số âm nào lớn hơn thì bé hơn, bé hơn thì lớn hơn
\(\Rightarrow\)\(-243>-125000\)
\(\Rightarrow\)\(\left(-3\right)^{50}>\left(-50\right)^{30}\)