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Viết các kết quả sau dưới dạng một lũy thừa: 

9¹². 27⁴. 81³ và 4⁸ . 2¹⁰ 

Giúp mk vs

6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7

7: =(1/2)^7*2^3*2^5*2^8=2^9

8: =(-1/7)^4*5^4=(-5/7)^4

9: =2^2*2^5:(2^3/2^4)

=2^7/2=2^6

10: =(1/7)^3*7^2=1/7

1) a) 4⁸.2²⁰ = (2²)⁸.2²⁰

= 2¹⁶.2²⁰ = 2³⁶

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9¹².27⁵.81³ = (3²)¹².(3³)⁵.(3⁴)⁴

= 3²⁴.3¹⁵.3¹⁶ = 3⁵⁵

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64³.4⁵.16² = (4³)³.4⁵.(4²)²

= 4⁹.4⁵.4⁴ = 4¹⁸

b) 25²⁰.125⁴ = (5²)²⁰.(5³)⁴

= 5⁴⁰.5¹² = 5⁵²

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x⁷.x³.x⁴ = x¹⁴

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3⁶.4⁶ = (3.4)⁶ = 12⁶

2) a) 2² = 4

2³ = 8

2⁴ = 16

2⁵ = 32

2⁶ = 64

2⁷ = 128

2⁸ = 256

2⁹ = 512

2¹⁰ = 1024

b) 3² = 9

3³ = 27

3⁴ = 81

3⁵ = 243

c) 4² = 16

4³ = 64

4⁴ = 256

d) 5² = 25

5³ = 125

5⁴ = 625

3)

a) 4⁹ : 4⁴ = 4⁵

17⁸ : 17⁵ = 17³

2¹⁰ : 8² = 2¹⁰ : (2³)² = 2¹⁰ : 2⁶ = 2⁴

18¹⁰ : 3¹⁰ = (18 : 3)¹⁰ = 6¹⁰

27⁵ : 81³ = (3³)⁵ : (3⁴)³ = 3¹⁵ : 3¹² = 3³

b) 10⁶ : 100 = 10⁶ : 10² = 10⁴

5⁹ : 25³ = 5⁹ : (5²)³ = 5⁹ : 5⁶ = 5³

4¹⁰ : 64³ = 4¹⁰ : (4³)³ = 4¹⁰ : 4⁹ = 4

2²⁵ : 32⁴ = 2²⁵ : (2⁵)⁴ = 2²⁵ : 2²⁰ = 2⁵

18⁴ : 9⁴ = (18 : 9)⁴ = 2⁴

Bài 1 :

a/   \(a^3.a^9=a^{3+9}=a^{12}\)

b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)

c/  \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)

d/  \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)

e/  \(5^6:5^3+3^3.3^2\)

     \(=5^3+3^5=125+243=368\)

i/ \(4.5^2-2.3^2\)

   \(=2^2.5^2-2.3^2\)

   \(=2^2.25-2^2.14\)

   \(=2^2.\left(25-14\right)\)

   \(=2^2.11\)      

   \(=4.11=44\)

Bài 2 :

a, \(3^8:3^6=3^{8-6}=3^2\)

 \(7^5:7^2=7^{5-2}=7^3\)

 \(19^7:19^3=19^{7-3}=19^4\)

  \(2^{10}.8^3=2^{10}.\left(2^3\right)^3=2^{10}.2^9=2^{19}\)  

\(12^7:6^7=\left(12:6\right)^7=2^7=128\)

\(27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3=9\)

Bạn Võ thị khánh ly ơi có thể giải giúp mk phần b bài 2 k

a: ${}^{}\cdot {}^{}={}^{}={}^{}$

b: 64⋅83=24⋅34⋅29=213⋅3464⋅83=24⋅34⋅29=213⋅34

c: $16\cdot 81={}^{}$

d: ${}^{}\cdot {}^{}={}^{}$

thu gọn 7^3*7^5

Oh no nhiều kí tự đặc biệt quá

dễ quá mình ko làm  đc 

\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)

\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)

\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)

\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
 

6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)

7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)

8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)

9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)

10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)

6) \(...=\dfrac{3}{2}.\left(\dfrac{3}{2}\right)^2.\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

7) \(...=2^{-7}.2^3.2^5.2^8=2^9\)

8) \(...=\left(\dfrac{1}{7}\right)^4.5^3.5=\left(\dfrac{1}{7}\right)^4.5^4\)

9) \(...=2^2.2^5:\left(2^3.2^{-4}\right)=2^8\)

10) \(...=\left(\dfrac{1}{7}\right)^3.\left(\dfrac{1}{7}\right)^2=\left(\dfrac{1}{7}\right)^5\)

\(a,a^{\dfrac{1}{3}}\cdot\sqrt{a}=a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}=a^{\dfrac{5}{6}}\\ b,b^{\dfrac{1}{2}}\cdot b^{\dfrac{1}{3}}\cdot\sqrt[6]{b}=b^{\dfrac{1}{2}}\cdot b^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{6}}=b^1\)

\(c,a^{\dfrac{4}{3}}:\sqrt[3]{a}=a^{\dfrac{4}{3}}:a^{\dfrac{1}{3}}=a^{\dfrac{4}{3}-\dfrac{1}{3}}=a\\ d,\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}=\sqrt[6]{b}\)