Lời giải:

$2520.1254=3160080^1$

a) \(...=10^5.10^2.10^2=10^9\)

b) \(...=\left(2^2.3\right).\left(2^2.3\right).\left(2^2.3\right).2.3=2^7.3^4\)

c) \(...=5^2.5.2^2.2.2.5=2^4.5^4\)

d) \(...=7.3.2.5.2.5.3.5.2.5=2^33^2.5^4.7\)

a/

\(100000.100.10.10=10^5.10^2.10.10=10^9=2^9.5^9\)

b/

\(12.12.12.6=2^2.3.2^2.3.2^2.3.2.3=2^7.3^4\)

c/

\(25.5.4.2.10=5^2.5.2^2.2.2.5=2^4.5^4\)

d/

\(210.10.3.5.10=21.10.10.3.5.10=3.7.2.5.2.5.3.5.2.5=2^3.3^2.5^4.7\)

Bài 6: 

a: \(2^{27}=8^9\)

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

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

a: \(3\cdot3\cdot3\cdot3\cdot3=3^5\)

b: \(y\cdot y\cdot y\cdot y=y^4\)

c: \(5\cdot p\cdot5\cdot p\cdot2\cdot q\cdot4\cdot q=25\cdot2\cdot4\cdot p^2q^2=2\cdot\left(10qp\right)^2\)

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

a) 3.3.3.3.3 = 3\(^5\)

b) y.y.y.y = y\(^4\)

c) 5.p.5.p.2.q.4.q = 5\(^2\).p\(^2\).q\(^2\).2\(^3\) 

(2\(^3\) ở đây là vì 2.4 = 2.2.2 = 2\(^3\))

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

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

a: \(A=2\cdot2^2\cdot...\cdot2^{10}\)

=>\(A=2^{1+2+...+10}\)

=>\(A=2^{55}\)

b: \(B=3\cdot3^3\cdot3^5\cdot...\cdot3^{99}\)

\(=3^{1+3+5+...+99}\)

\(=3^{50^2}=3^{2500}\)

a) 2.4.8.8 = 2.2.2.2.2.2.2.2.2 = 2 9

-4^3.(-5)^2

4^3.4^7 = 4^3+7 = 4^10

ta có :

\(3^3.7^3=\left(3.7\right)^3=21^3\)

3³7³=(3.7)³