5.p.p.5.p.2.q.q = 5².p³.q².2 

\(5.p.p.5.p^2.q.4.q=\left(5.5.4\right).\left(p.p.p^2\right).\left(q.q\right)=100p^4.q^2\)

5.p.p.5.p².q.4.q

= 5².p⁴.2².q²

= 10².p⁴.q²

Ta có: 712 : 74 = 7(12-4) = 78

(x^m)ⁿ

( x ^m ) ⁿ

Ta có: a4 : a4 = a(4-4) = a0

x6 : x3 = x(6-3) = x3

a) 5³ . 5² . 5 

= 5⁵ . 5 

= 5⁵ . 5¹

= 5⁶

b) 6⁹ : 6⁴ 

= 6⁵

c) 4 . 8 . 16 . 32 

= 2² . 2³ . 2⁴ . 2⁵

= 2⁵ . 2⁴ . 2⁵

= 2⁹ . 2⁵

= 2¹⁴ 

Bài 6: 

a: \(2^{27}=8^9\)

\(3^{18}=9^9\)

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

\(2^7\cdot2^9\cdot3^2=2^{7+9}\cdot3^2=2^{16}\cdot3^2\)

chỉ bt chừng đó thôi 😅

2⁷.2⁹.3² = 2⁷⁺⁹.9 = 2¹⁶.9 = 65536.9 = 589824

í, sửa lại 1 tíu

\(=65536\cdot9=589824\)

\(a,5\cdot p\cdot5\cdot p\cdot2q\cdot4q\)

\(=5^2\cdot p^2\cdot2\cdot q\cdot2^2\cdot q\)

\(=5^2\cdot2^3\cdot p^2\cdot q^2\)

\(b,a\cdot a+b\cdot b+c\cdot c\cdot c+d\cdot d\cdot d\cdot d\)

\(=a^2+b^2+c^3+d^4\)

#Urushi