a, 2^24 > 3^16

b, 5^300>3 ^500

c,99^20 > 9999^10

d, 2^30 +3^44 +4^30 < 3x24^10

Câu a:

2\(^{300}\) và 3\(^{200}\)

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

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

8\(^{100}\) < 9\(^{100}\)

Vậy 2\(^{300}\) < 3\(^{200}\)

câu b:

99\(^{20}\) và 9999\(^{10}\)

99\(^{20}\) = (99\(^2\))\(^{10}\) = 9801\(^{10}\)

9999\(^{10}\) > 9801\(^{10}\)

Vậy 99\(^{20}\) < 9999\(^{10}\)

Câu c:

3\(^{500}\) và \(7^{300}\)

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

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

243\(^{100}\) < 343\(^{100}\)

Vậy 3\(^{500}\) < 7\(^{300}\)

Câu d:

11\(^{1979}\) và 37\(^{1320}\)

11\(^{1979}\) < 11\(^{1980}\) = (11\(^3\))\(^{660}\) = 1331\(^{660}\)

37\(^{1320}\) = (37\(^2\))\(^{660}\) = 1369\(^{660}\)

1331\(^{660}<1369^{660}\)

Vậy 11\(^{1979}\) < 37\(^{1320}\)

Câu e:

so sánh : 8\(^5\) và 3.4\(^7\)

8\(^5\) = (2\(^3\) )\(^5\) = 2\(^{15}\) = 2.(2\(^2\))\(^7\)= 2.4\(^7\)

3.4\(^7\) > 2.4\(^7\)

Vậy: 8\(^5\) < 3.4\(^7\)

Câu f; So sánh:

10\(^{10}\) và 48.50\(^5\)

48.50\(^5\) = (2\(^4\).3).50\(^5\) > (2\(^4.2\)).50\(^5\) = 2\(^5\).50\(^5\) = (2.50)\(^5\)=100\(^5\) =10\(^{10}\)

Vậy 10\(^{10}\) < 48.50\(^5\)

1/2^1050 <   5^450

2/5^2n   < 2^2n

3/3^500  <  7^300

4/8^5 <   3.4^7

5/99^20 < 9999^10

k mk nha

giai han ra ban ak

Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

\(d,10^{20}=\left(10^2\right)^{10}=100^{10}\\ Vì:100^{10}>20^{10}\left(Do:100>20\right)\\ \Rightarrow10^{20}>20^{10}\)

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

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

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

\(b,8^5=32768\)

\(6^6=46656\)

Vì \(32768< 46656\) nên \(8^5< 6^6\)

\(c,3^{450}=\left(3^3\right)^{150}=27^{150}\)

\(5^{300}=\left(5^2\right)^{150}=25^{150}\)

Vì \(27^{150}>25^{150}\) nên \(3^{450}>5^{300}\)

#Ayumu

f: 11^1979<11^1980=1331^660

37^1320=(37^2)^660=1369^660

1331<1369

=>1331^660<1369^660

=>11^1980<37^1320

=>11^1979<37^1320

g: 10^10=2^10*5^10

48*50^5=2^4*3*2^5*5^10=2^9*3*5^10

2^10<2^9*3

=>2^10*5^10<2^9*3*5^10

=>10^10<48*50^5

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!