5³⁶=(5³)¹²=125¹²

11²⁴=(11²)¹²=121¹²

125¹².121¹²=>5³⁶>11²⁴

Vậy 5³⁶>11²⁴

a) \(a^9:a^7\cdot\left(a^2\right)^2\)

\(=a^9:a^7\cdot a^4\)

\(=a^2\cdot a^4\)

\(=a^6\)

b) \(\left(ab\right)^6:b^5:b\)

\(=a^6b^6:b^4\)

\(=a^6b^2\)

thanks

a) \(9x^2-12x+4=\left(3x-2\right)^2\)

b) \(25+10x+x^2=\left(x+5\right)^2\)

c) \(36x^2-25=\left(6x-5\right)\left(6x+5\right)\)

d) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\)

a: \(9x^2-12x+4=\left(3x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

c: \(36x^2-25=\left(6x-5\right)\left(6x+5\right)\)

5^300 on hon 36453

a) \(4^9:4^4\)

\(=4^{9-4}\)

\(=4^5\)

b) \(17^8:17^5\)

\(=17^{8-5}\)

\(=17^3\)

c) \(2^{10}:8^2\)

\(=2^{10}:\left(2^3\right)^2\)

\(=2^{10-6}\)

\(=2^4\)

d) \(18^{10}:3^{10}\)

\(=\left(18:3\right)^{10}\)

\(=6^{10}\)

e) \(27^5:81^3\)

\(=\left(3^3\right)^5:\left(3^4\right)^3\)

\(=3^{15}:3^{12}\)

\(=3^{15-12}\)

\(=3^3\)

f) \(10^6:100\)

\(=10^6:10^2\)

\(=10^{6-2}\)

\(=10^4\)

g) \(5^9:25^3\)

\(=5^9:\left(5^2\right)^3\)

\(=5^9:5^6\)

\(=5^{9-6}\)

\(=5^3\)

h) \(4^{10}:64^3\)

\(=4^{10}:\left(4^3\right)^3\)

\(=4^{10-9}\)

\(=4\)

i) \(2^{25}:32^4:18^4:9^4\)

\(=\left(2^{25}:2^{20}\right):\left(18^4\cdot9^4\right)\)

\(=2^5:9^8:2^4\)

\(=2:9^8\)

\(=\dfrac{2}{9^8}\)

a,4^9-4

b,17^8-5

c,2^10-5

d,6^10

e,81

f,

8^5<6^6

32.....<46.......

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

3^500=(3^5)^100=243^100

Vì 125<243 nên 125^100<243^100

Vậy 125^100<243^100

tk cho mk nha bn

\(5^{300}=5^{3.100}=125^{100}\)

\(3^{500}=3^{5.100}=243^{100}\)

vì 125<243 nên \(5^{300}< 3^{500}\)