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) \({2^m}{.2^n}=\underbrace {2.2 \ldots .2}_{m{\rm{ }}}{\rm{ }}.\underbrace {2.2 \ldots .2}_{n{\rm{ }}}{\rm{ }}\) = 2^m+n

b) \({3^m}:{3^n}=(\underbrace {3.3 \ldots .3}_{m{\rm{ }}}{\rm{ }}):(\underbrace {3.3 \ldots .3}_{n{\rm{ }}}{\rm{ }})\) = 3^m-n  với \(m \ge n\)

a. \(3^4.3^5=3^9\)

\(16.2^9=2^4.2^9=2^{13}\)

\(16.32=2^4.2^5=2^9\)

b. \(12^8.12=12^9\)

\(243:3^4=3^5:3^4=3\)

\(10^9:10000=10^9:10^4=10^5\)

c. \(4.8^6.2.8^3=2^2.2^{18}.2.2^9=2^{30}\)

\(12^2.2.12^3.6=12^2.12.12^3=12^6\)

\(6^3.2.6^4.3=6^3.6.6^4=6^8\)

he

a. 

16.29=24.29=21316.29=24.29=213

b. 

c. 4.86.2.83=22.218.2.29=2304.86.2.83=22.218.2.29=230

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`

a) 3.3^4.3^5 = 3^10

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

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

Chọn B

B 

=> \(^{2^8}\) = 256

B

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: 

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

2^(10 : 64 x 16) = 2^[(10 x 16) : 64] = 2^(160 : 64) = 2^25

2^(10 x 8) = 2^80     3^(5 : 27) = 3^(5/27)     5^(2 x 125) = 5^250     6^(6 : 36) = 6^(1/6)

a)\({\left[ {{{\left( { - \frac{1}{6}} \right)}^3}} \right]^4}\) (với \(a =  - \frac{1}{6}\))

\(=(- \frac{1}{6})^{3. 4}=(- \frac{1}{6})^{12}\)

b)\({\left[ {{{\left( { - 0,2} \right)}^4}} \right]^5}\) (với \(a =  - 0,2\))

\(=(-0,2)^{4.5}=(-0,2)^{20}\)