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

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\(^4\)

\(\begin{array}{l}a){( - 2)^3}.{( - 2)^4} = {( - 2)^{3 + 4}} = {( - 2)^7}\\b){(0,25)^7}:{(0,25)^3} = {(0,25)^{7 - 3}} = {(0,25)^4}\end{array}\)

a) 3³ . 3⁴ = 3⁷

b) 5² . 5⁷ = 5⁹

c) 7⁵ . 7 = 7⁶

a) \(3^3.3^4=3^{3+4}=3^7\)

b) \(5^2.5^7=5^{2+7}=5^9\)

c) \(7^5.7=7^{5+1}=7^6\)

\(3^3.3^4=3^{3+4}=3^7\)

\(5^2.5^7=5^{2+7}=5^9\)

\(7^5+7=7^{5+1}=7^6\)

Sao chẳng có ai T mik z bất công quá

a. 2^3 . 2^2 . 2^4

= 2^9

b. 10^2 . 10^3 . 10^5

= 10^10

c. x. x^5

= x^6

d. a^3 . a^2 . a^5

= a^10

a) \(2^3.2^2.2^4\)

   \(=2^{3+2+4}\)

   \(=2^9\)

b) \(10^2.10^3.10^5\)

   \(=10^{2+3+5}\)

   \(=10^{10}\)

c) \(x.x^5=x^1.x^5=x^6\)

d) \(a^3.a^2.a^5=a^{3+2+5}=a^{10}\)

2⁵ . 4³ = 2⁵ . (2²)³ = 2⁵ . 2⁶ = 2^{5 + 6} = 2¹¹

\(2^{11}\)

\(2^5\cdot4^3=2^5\cdot2^6=2^{11}\)

Bài 6: 

a: \(2^{27}=8^9\)

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

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)