1. 5³ = 5.5.5 = 125

2. 2⁷ = 2.2.2.2.2.2.2 = 128

3. 4⁴ = 4.4.4.4 = 256

4. 7³ = 7.7.7 = 343

6. 3⁵ = 243

7. 2⁶ =  64

8. 3⁴ =  81

9. 8³ =  512

11. 13² = 169

12. 11² = 121

13. 14² = 196

14. 15² = 225

16. 17² = 289

17. 18² = 324

18. 19² = 361

19. 20² = 400

21. 10⁴ = 10000

22. 10⁵ = 100000

23. 10⁶ = 1000000

24. 10⁷ = 10000000

bạn làm như bạn gia hân là đúng nhé

giải giúp mình bài 2 trên với ạ

a,\(5^3.2-100:4+2^3.5\)

= 125 . 2 - 25 + 8 . 5

= 250 - 25 + 40

= 265

b, \(6^2:9+50.2-3^3.3\)

= 36 : 9 + 100 - 27 . 3

= 4 + 100 - 81

= 23

bạn kia ko làm các ý khác à

b) \(5^3\cdot2-100:4+2^3\cdot5\)

\(=125\cdot2-25+8\cdot5\)

\(=250-25+40\)

\(=225+40=265\)

c) \(6^2:9+50\cdot2+3^3-3\)

\(=36:9+100+27-3\)

\(=4+100+27-3\)

\(=104+27-3=131-3=128\)

d) \(3^2\cdot5+2^3\cdot10-81:3\)

\(=9\cdot5+8\cdot10-27\)

\(=45+80-27\)

\(=125-27=98\)

e) \(5^{13}:5^{10}-25\cdot2^2\)

\(=5^{13-10}-5^2\cdot2^2\)

\(=5^3-\left(5\cdot2\right)^2\)

\(=125-10^2\)

\(=125-100=25\)

f) \(20:2^2+5^9:5^8\)

\(=20:4+5^{9-8}\)

\(=5+5^1=5+5=10\)

g) \(100:5^2+7\cdot3^2\)

\(=10^2:5^2+7\cdot9\)

\(=\left(10:5\right)^2+63\)

\(=2^2+63=4+63=67\)

h) \(84:4+3^9:3^7+5^0\)

\(=21+3^{9-7}+1\)

\(=21+3^2+1\)

\(=21+9+1=30+1=31\)

i) \(29-\left[16+3\cdot\left(51-49\right)\right]\)

\(=29-\left[16+3\cdot2\right]\)

\(=29-\left[16+6\right]\)

\(=29-22=7\)

j) \(\left(15^{19}:5^{17}+3\right)\cdot0:7\)

\(=\left[\left(3\cdot5\right)^{19}:5^{17}+3\right]\cdot0\)

Vì số nào nhân cho 0 cũng bằng 0 nên giá trị biểu thức trên bằng 0

k) \(7^9:7^7-3^2+2^3\cdot5\)

\(=7^{9-7}-9+8\cdot5\)

\(=7^2-9+40\)

\(=49-9+40=40+40=80\)

l) \(1200:2+6^2\cdot2^1+18\)

\(=600+36\cdot2+18\)

\(=600+72+18\)

\(=600+\left(72+18\right)=600+90=690\)

m) \(5^9:5^7+70:14-20\)

\(=5^{9-7}+5-20\)

\(=5^2+5-20\)

\(25+5-20=30-20=10\)

Những câu sau mình làm sau nhé bạn!!!!!!!

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Bài 3

2\(^x\) = 16

2\(^x\) = 2⁴

  \(x=4\)

Vậy \(x=4\)

a) 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

c) 6² = 36; 6³ = 216; 6⁴ = 1296

a) 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

e) 6² = 36; 6³ = 216; 6⁴ = 1296

HT

4 mũ 3- 2×(4- 2 mũ3)

\(a,\)\(\frac{2^5\times3^{12}\times7^8}{2^7\times3^{10}\times7^9}=\frac{3^2\times\left(2^5\times3^{10}\times7^8\right)}{2^2\times7\times\left(2^5\times3^{10}\times7^8\right)}\)\(=\frac{3^2}{2^2\times7}=\frac{9}{28}\)

\(b,\)Tương tự

1; 7³.5².5⁴.7⁶:(5⁵.7⁸)

= (7³.7⁶).(5².5⁴) : (5⁵.7⁸)

= 7⁹.5⁶: (5⁵.7⁸)

= (7⁹:7⁸).(5⁶:5⁵)

= 7.5

= 35

2; 3³.a⁷.3.a²:(3⁴.a⁶)

= (3³.3).(a⁷.a²): (3⁴.a⁶)

= 3⁴.a⁹: (3⁴.a⁶)

= (3⁴:3⁴).(a⁹:a⁶)

= a³

3; 7³.11⁴.a⁸.b⁷: 7².11².a⁵.b⁶

=  (7³:7²).(11⁴.11²).(a⁸.a⁵).(b⁷.b⁶)

= 7.11⁶.a¹³.b¹³

a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)

 1/2 x 2 mũ n cộng 4 x 2 mũ n = 9 x 2 mũ n

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)