a) 3.3^4.3^5 = 3^10

b) 7^3 : 7^2 : 7 = 7^1

c) (x^4)^3 = x^12

a)  \(\frac{6}{5}.{\left( {1,2} \right)^8} = 1,2.{(1,2)^8} = {(1,2)^{1 + 8}} = {(1,2)^9}\)

b) \({\left( {\frac{{ - 4}}{9}} \right)^7}:\frac{{16}}{{81}} = {\left( {\frac{{ - 4}}{9}} \right)^7}:{\left( {\frac{{ - 4}}{9}} \right)^2} = {\left( {\frac{{ - 4}}{9}} \right)^{7 - 2}} = {\left( {\frac{{ - 4}}{9}} \right)^5}\)

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) \(3^3 : 3^2\)=\(3^{3-2}=3^1\)

b) \(5^{4} : 5^{2}\)=\(5^{4-2}=5^2\)

c) \(8^{3} . 8^{2}\)=\(8^{3+2}=8^5\)

d) Cách 1:

\(5^{4} . 5^{3}: 5^{2}=5^{4+3}:5^2=5^7:5^2=5^{7-2}=5^5\)

Cách 2: 

\(5^{4} . 5^{3}: 5^{2}=5^{4+3-2}=5^5\)

Chọn B

B 

=> \(^{2^8}\) = 256

B

a) \(5^{20}.5=5^{20+1}=5^{21}\)

b) \(7^9:7=7^{9-1}=7^8\)

a)5²⁰ x 5 = 5²⁰ x 5¹ =5⁽²⁰⁺¹⁾=5²¹

b)7⁹ : 7 = 7⁹ : 7¹ = 7^(9-1) = 7⁸

a)  5²⁰ x 5 = 5^{20 +1} = 5²¹

b) 7⁹ : 7 = 7^9-1 = 7⁸

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

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

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

a,3³.3⁴=3⁷

b,5².5⁷=5⁹

c,7⁵.7=7⁶

a)    3³.3⁴=3⁷

b)     5².5⁷=5⁹

c)     7⁵.7=7⁶

Ai kick mình mình kick lại

a) 3³ . 3⁴ = 3⁷

b) 5² . 5⁷ = 5⁹

c) 7⁵ . 7 = 7⁶

a, 3³ . 3⁴ = 3⁷

b, 5² . 5⁷ = 5⁹

c, 7⁵ . 7 = 7⁶

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

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

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

a) 3^7=2187

b) 5^9=1953125

c) 7^6=117649

a     3^7=2187

b 5^9=9153125

c    7^6=117649

a) `7^4 .7^5 .7^6 = 7^(4+5+6)=7^(15)`

b) `(54:3^7) .324 =( 2.3^3 : 3^7).2^2 x3^4`

`=(2: 3^4).3^4 .2^2`

`= 2.2^2=2^3`

c) `[(8+2^2).10^100] : (10^0 .10^94)`

`=(12 . 10^100) : 10^94`

`=12 . (10^100 : 10^94)`

`=12 . 10^6`

d) `a^9 :a^9 = a^(9-9)=a^0`