1+2^2+3^2+4^2+5^2+...+99^2

125²⁰=(5³)²⁰=5⁶⁰

256¹⁵=(4⁴)¹⁵=4⁶⁰

5⁶⁰ > 4⁶⁰ . Vậy 125²⁰ > 256¹⁵

Ta có:

\(125^{20}=\left(5^3\right)^{20}=5^{60}\)

\(256^{15}=\left(4^4\right)^{15}=4^{60}\)

Vì \(5^{60}>4^{60}\) nên \(125^{20}>256^{15}.\)

Ta có: \(125^{20}=\left(5^3\right)^{20}=5^{60}\)

\(256^{15}=\left(4^4\right)^{15}=4^{60}\)

Từ đó suy ra \(5^{60}>4^{60}\) Hay \(125^{20}>256^{15}\)(đpcm)

a) 125⁸⁰=Math ERROR
342.348=119016<Math ERROR
b) 10³⁰=(10³)¹⁰=1000¹⁰<1000²⁰

a)9^20 và 27^13

9^20=(3^2)^20=3^40

27^13=(3^3)^13=3^39

vì 3^40 > 3^39 =>9^20>27^13

b)10^30 và 2^100

10^30=(10^3)^10=30^10

2^100=(2^10)^10=20^10

vì 30^10>20^0 => 10^30>2^100

c)125^5 và 25^7

125^5=(5^3)^5=5^15

25^7=(5^2)^7=5^14

vì 5^15>5^14 =>125^5>25^7

Ta có :

a) \(9^{20}=\left(3^2\right)^{20}=3^{40};27^{13}=\left(3^3\right)^{13}=3^{39}\)

Vì \(3^{40}>3^{39}\Rightarrow9^{20}>27^{13}\)

Vậy \(9^{20}>27^{13}\)

a, \(125^{20}\)và \(25^{30}\)

ta có : \(125^{20}=\left(5^3\right)^{20}\)\(=5^{3.20}=5^{60}\)

\(25^{30}=\left(5^2\right)^{30}=5^{2.30}=5^{60}\)

Vì \(5^{60}=5^{60}\)nên => \(125^{20}=25^{30}\)

b ,\(49^{16}\)và \(343^{20}\)

ta có : \(49^{16}=\left(7^2\right)^{16}=7^{2.16}=7^{32}\)

\(343^{20}=\left(7^3\right)^{20}=7^{3.20}=7^{60}\)

Vì \(7^{32}< 7^{60}\)nên => \(49^{16}< 343^{20}\)

c, \(121^{15}\)và \(1331^{16}\)

ta có : \(121^{15}=\left(11^2\right)^{15}=11^{2.15}=11^{30}\)

\(1331^{16}=\left(11^3\right)^{16}=11^{3.16}=11^{48}\)

Vì \(11^{30}< 11^{48}\)nên => \(121^{15}< 1331^{16}\)

d, \(199^{20}\)và \(2003^{15}\)

ta có : \(199^{20}=199^{5.4}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=2003^{3.5}=\left(2003^3\right)^5=8036054027^5\)

Vì \(1568239201^5< 8036054027^5\)nên => \(199^{20}< 2003^{15}\)

e, \(4^{25}\)và \(3^{30}\)

=> \(4^{25}< 3^{30}\)

f, \(36^{82}\)và \(49^{123}\)

=> \(36^{82}< 49^{123}\)

mình làm rồi đó . k mình đi

cái nào mũ lớn hơn thì nó lớn lơn

a)Ta có : \(3^{39}=\left(3^{13}\right)^3=1594323^3\)
  \(11^{21}=\left(11^7\right)^3=19487171^3\)
Vì \(1594323< 19487171\)
\(=>1594323^3< 19487171^3\)
\(=>3^{39}< 11^{21}\)
Vậy \(3^{39}< 11^{21}\)
   b)Ta có : \(199^{20}=\left(199^4\right)^5=1568239201^5\)
                \(2002^{15}=\left(2002^3\right)^5=8024024008^5\)
Vì \(1568239201< 8024024008\)
\(=>1568239201^5< 8024024008^5\)
\(=>199^{20}< 2002^{15}\)
Vậy \(199^{20}< 2002^{15}\)
c) Ta có:\(125^{90}=\left(125^3\right)^{30}=1953125^{30}\)
            \(25^{120}=\left(25^4\right)^{30}=390625^{30}\)
Vì \(1953125>390625\)
\(=>1953125^{30}>390625^{30}\)
\(=>125^{90}>25^{120}\)
Vậy \(125^{90}>25^{120}\)
d)Ta có : \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
              \(7^{300}=\left(7^3\right)^{100}=343^{100}\)
Vì \(243< 343\)
\(=>243^{100}< 343^{100}\)
\(=>3^{500}< 7^{300}\)
Vậy \(3^{500}< 7^{300}\)
Phù , cuối cùng cũng viết xong . Mỏi tay quá ! À , chúc bạn học tốt nhé !

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a,

15^12=(3*5)^12=3^12*5^12

81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15

Vì 12<15 suy ra 5^12<5^15

Suy ra 3^12*5^12<3^12*5^15

\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)

\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)

\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)

Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)