Bài 1:

\(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{25^2}<\frac{1}{24\cdot25}=\frac{1}{24}-\frac{1}{25}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{25^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{24}-\frac{1}{25}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{25^2}<1-\frac{1}{25}<1\)

a: \(\frac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(=\frac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}=\frac{2^8\cdot5^7\left(2^3+3^3\right)}{2^5\cdot5^4\left(2^3+3^3\right)}\)

\(=2^3\cdot5^3=10^3=1000\)

b: \(149-\left(35:x+3\right)\cdot17=13\)

=>\(\left(35:x+3\right)\cdot17=149-13=136\)

=>35:x+3=136:17=8

=>35:x=8-3=5

=>\(x=\frac{35}{5}=7\)

c: \(121:11-\left(4x+5:3\right)=4\)

=>11-(4x+5/3)=4

=>4x+5/3=11-4=7

=>\(4x=7-\frac53=\frac{21}{3}-\frac53=\frac{16}{3}\)

=>\(x=\frac{16}{3}:4=\frac43\)

d: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=40\)

=>\(41-\left(2x-5\right)=\frac{720}{40}=18\)

=>2x-5=41-18=23

=>2x=28

=>\(x=\frac{28}{2}=14\)

e: Sửa đề: (x+1)+(x+2)+(x+3)+...+(x+100)=5700

=>100x+(1+2+3+...+100)=5700

=>\(100x+100\cdot\frac{101}{2}=5700\)

=>x+50,5=57

=>x=57-50,5

=>x=6,5

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

\(2^{x+1}-2^x=3^2\)

\(\Rightarrow2^x\cdot\left(2-1\right)=9\)

\(\Rightarrow2^x=9\)

\(\Rightarrow x\in\varnothing\)

\(2^{x+1}-2^x=3^2\)

=>2^x*2-2^x=9

=>2^x=9

=>\(x\in\varnothing\)

\(-\dfrac{1}{9}:\dfrac{13}{8}+\left(-\dfrac{1}{9}\right):\dfrac{13}{5}+\left(-\dfrac{2}{3}\right)^2\)
\(=\left(-\dfrac{1}{9}\right)\times\dfrac{8}{13}+\left(-\dfrac{1}{9}\right)\times\dfrac{5}{13}+\dfrac{4}{9}\)
\(=\left(-\dfrac{1}{9}\right)\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{4}{9}\)
\(=\left(-\dfrac{1}{9}\right)\cdot1+\dfrac{4}{9}\)
\(=\left(-\dfrac{1}{9}\right)+\dfrac{4}{9}\)
\(=\dfrac{3}{9}=\dfrac{1}{3}\)

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

2357tthhhh

A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

a)\(3^2.9^3=9.9^3=9^{1+3}=9^4\)

b)\(2^2.5^2=4.25=100=10^2\)

c)\(8^5.2^3=8^5.8=8^{5+1}=8^6\)

d)\(9^8:3^2=9^8:9=9^{8-1}=9^7\)

a)81 mũ 3

b)10 mũ 2

c)64 mũ  5

d)9 mũ 7

k cho mk nhé