a: \(4^8\cdot2^{20}=\left(2^2\right)^8\cdot2^{20}=2^{16}\cdot2^{20}=2^{16+20}=2^{36}\)

\(9^{12}\cdot27^5\cdot81^4=\left(3^2\right)^{12}\cdot\left(3^3\right)^5\cdot\left(3^4\right)^4\)

\(=3^{24}\cdot3^{15}\cdot3^{16}=3^{24+15+16}=3^{55}\)

\(64^3\cdot4^5\cdot16^2=\left(4^3\right)^3\cdot4^5\cdot\left(4^2\right)^2=4^9\cdot4^5\cdot4^4=4^{9+5+4}=4^{18}\)

b: \(25^{20}\cdot125^4=\left(5^2\right)^{20}\cdot\left(5^3\right)^4=5^{40}\cdot5^{12}=5^{52}\)

\(x^7\cdot x^4\cdot x^3=x^{7+4+3}=x^{14}\)

\(3^6\cdot4^6=\left(3\cdot4\right)^6=12^6\)

c: \(8^4\cdot2^3\cdot16^2=\left(2^3\right)^4\cdot2^3\cdot\left(2^4\right)^2\)

\(=2^{12}\cdot2^3\cdot2^8=2^{23}\)

\(y\cdot y^7=y^{1+7}=y^8\)

\(2^3\cdot2^2\cdot8^3=2^5\cdot\left(2^3\right)^3=2^5\cdot2^9=2^{5+9}=2^{14}\)

Bài 1: 

a) \(4^8\cdot2^{20}=\left(2^2\right)^8\cdot2^{20}=2^{36}\)

 \(64^3\cdot4^5=\left(2^6\right)^3\cdot\left(2^2\right)^5=2^{18}\cdot2^{10}=2^{28}\)

\(y\cdot y^7=y^{1+7}=y^8\)

\(a^n\cdot a^2=a^{n+2}\)

Bài 1:

b) \(10^8:2^8=5^8\)

\(17^8:17^5=17^3\)

\(2^{25}:32^4=2^{25}:2^{20}=2^5\)

\(19^4:9^4=\left(\dfrac{19}{9}\right)^4\)

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 \(2^{27}< 3^{18}\)

cảm ơn nha

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

cảm ơn cậu nhiều

5.p.p.5.p.2.q.q = 5².p³.q².2 

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

a: \(A=2\cdot2^2\cdot...\cdot2^{10}\)

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

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

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

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

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

Anh ơi em on rồi ạ anh trả lời câu hỏi của em với nha anh thanks anh nhiều ạ

a) 2.4.8.8 = 2.2.2.2.2.2.2.2.2 = 2 9