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

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

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

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

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

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

#Urushi

\(\begin{array}{l}{3^3}{.3^4} = {3^{3 + 4}} = {3^7};\\{10^4}{.10^3}  = 10^{4+3}= {10^7};\\{x^2}.{x^5} = x^{2+5} = {x^7}.\end{array}\

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\(^4\)

l) 27^8=(3³)⁸=3²⁴

=>27⁸:3²²=3²⁴:3²²

=3^24-22=3²

m) 8^4=(2³)⁴=2¹²

4⁵=(2²)⁵=2¹⁰

=>8⁴:4⁵=2¹²:2¹⁰

=2^12-10=2²

a) \(27^8:3^{22}\)

\(=\left(3^3\right)^8:3^{22}\)

\(=3^{24}:3^{22}=3^{24-22}\)

\(=3^2\)

b) \(8^4:4^5\)

\(=\left(2^3\right)^4:\left(2^2\right)^5\)

\(=2^{12}:2^{10}=2^{12-10}=2^2\)

Anh đã từng làm em hi!

1.
a) \(3^4\times3^5\times3^6=3^{4+5+6}=3^{15}\)

b) \(5^2\times5^4\times5^5\times25=5^2\times5^4\times5^5\times5^2=5^{2+4+5+2}=5^{13}\)

c) \(10^8\div10^3=10^{8-3}=10^5\)

d) \(a^7\div a^2=a^{7-2}=a^5\)

2.

\(987=900+80+7\\ =9\times100+8\times10+7\\ =9\times10^2+8\times10^1+7\times10^0\)

\(2021=2000+20+1\\ =2\times1000+2\times10+1\times1\\ =2\times10^3+2\times10^1+1\times10^0\)

\(abcde=a\times10000+b\times1000+c\times100+d\times10+e\times1\\ =a\times10^4+b\times10^3+c\times10^2+d\times10^1+e\times10^0\)

còn bài 2