8 mũ 300=(8 mũ 3)mũ 100=512 mũ 100

9 mũ 200=(9 mũ 2) mũ 100=81 mũ 100

vì 512 mũ 100< 81 mũ 100 =>8 mũ 300 > 9 mũ 200

chắc chắn 100 %

Ta có :

\(8^{300}=\left(8^3\right)^{100}=512^{100}\)

\(9^{200}=\left(9^2\right)^{100}=91^{100}\)

Vì \(512^{100}>91^{100}\Rightarrow8^{300}>9^{200}\)

8³⁰⁰  va 9²⁰⁰

8³⁰⁰ = (8³)¹⁰⁰=512¹⁰⁰

9²⁰⁰  =(9²)¹⁰⁰=81¹⁰⁰

Vi 512>81 nen 8³⁰⁰>9²⁰⁰

\(10^{30}vs\)\(2^{100}\)

\(10^{30}=\left(10^3\right)^{10}=1000^{10}\)

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì \(1000^{10}< 1024^{10}=>10^{30}< 2^{100}\)

\(3^{54}vs2^{81}\)

\(3^{54}=\left(3^6\right)^9=729^9\)

\(2^{81}=\left(2^9\right)^9=512^9\)

Vì \(729^9>512^9=>3^{54}>2^{81}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(8^{100}< 9^{100}=>2^{300}< 3^{200}\)

a) Ta có:

\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)

Mà: \(8< 9\)

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) Ta có:

\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)

Mà: \(243< 343\)

\(\Rightarrow243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

c) Ta có: 

\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)

\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)

Mà: \(2< 3\)

\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)

\(\Rightarrow8^5< 3\cdot4^7\)

d) Ta có:

\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)

Mà: \(8242408>91809\)

\(\Rightarrow8242408^{101}>91809^{101}\)

\(\Rightarrow202^{303}>303^{202}\)

a ) 2²⁰⁰<8³⁰⁰

b ) 25²⁰⁰>5³⁰⁰

c ) 4²¹<64⁷

a)\(8^{300}=\left(2^3\right)^{300}=2^{900}\)

Vì \(200< 900\Rightarrow2^{200}< 8^{300}\)

b)\(25^{200}=\left(5^2\right)^{200}=5^{400}\)

Vì \(400>300\Rightarrow25^{200}>5^{300}\)

c)\(64^7=\left(4^3\right)^7=4^{21}\)

Vì \(4^{21}=4^{21}\Rightarrow4^{21}=64^7\)

a) \(2^{200}=2^{200}\)

\(8^{300}=\left(2^3\right)^{300}=2^{900}\)

vi \(2^{200}< 2^{900}\)nen \(2^{200}< 8^{300}\)

b) \(25^{200}=\left(5^2\right)^{200}=5^{400}\)

\(5^{300}=5^{300}\)

vi \(5^{400}>5^{300}\)nen \(25^{200}>5^{300}\)

c) \(4^{21}=\left(2^2\right)^{21}=2^{42}\)

\(64^7=\left(2^6\right)^7=2^{42}\)

vi \(2^{42}=2^{42}\)nen \(4^{21}=64^7\)

rgergqrgqrg

rgerger

\(5^{200}=25^{100}\)

\(3^{300}=27^{100}\)

mà 25<27

nên \(5^{200}< 3^{300}\)

Theo đề bài ta có :

\(n^{200}< 5^{300}\)( với n lớn nhất )

\(\left(n^2\right)^{100}< \left(5^3\right)^{100}\)

\(\left(n^2\right)^{100}< 125^{100}\)

\(n^2< 125\)

\(\Rightarrow n^2\in\left\{0;1;2;...;124\right\}\)

mà n lớn nhất \(\Rightarrow n^2=124\)

\(\Rightarrow n=\sqrt{124}\)

ta co 5^300=(5^3)^100=125^100

         n^200=(n^2)^100

nen n^2<125 suy ra n=11

trả lời tiếp :

mà n nguyên => n^2 phải có nghiệm nguyên

=> n^2 = 121

=> n = 11

Vậy số nguyên n lớn nhất thỏa mãn là 11

so sánh à?

a)\(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

\(2^{300}< 3^{200}\)

b)\(125^5=\left(25.5\right)^5=\left(5.5.5\right)^5=5^{15}\)

\(25^7=\left(5.5\right)^7=5^{14}\)

\(125^5>25^7\)

c)\(9^{20}=\left(3.3\right)^{20}=3^{40}\)

\(27^{13}=\left(3.3.3\right)^{13}=3^{39}\)

\(9^{20}>27^{13}\)

(^) là mũ

a)2^300 và 3^200

2^300=(2^3)^100=8^100

3^200=(3^2)^100=9^100

9^100>8^100

=>3^200>2^300

B)125^5 và 25^7

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

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

5^15>5^14

=>125^5>25^7

C)9^20 VÀ 27^13

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

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

3^39>3^40

=>27^13>9^20

CHÚC BẠN HỌC TỐT:D

:D

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