1+2+3+4+5+6+7+8+9+10=55

11+12+13+14+15+16+17+18+19+20=155

1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362

1+2+3+4+5+6+7+8+9+10=55

11+12+13+14+15+16+17+18+19+20=155

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362

\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

Biến đổi tử số 

\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)

= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)

= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)

= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)

Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)

Vậy A = 20

c.ơn nhìu a

a. ( 23 - 21) + ( 19 - 17) + ( 15 - 13) + ( 11 - 9) + ( 7 - 5) + ( 3 - 1) 

= 2 + 2 + 2 + 2 + 2 + 2 

= 2 x 6

= 12

b. ( 24 - 22 ) + ( 20 - 18 ) + ( 16 - 14 ) + ( 12 - 10) + ( 8 - 6 ) + ( 4 - 2) 

= 2 + 2 + 2 + 2 + 2 + 2 

= 2 x 6 

= 12

Giải:

a) Sửa đề: 127² + 146.127 + 73²

\(127^2+146.127+73^2=\left(127+7\right)^2=200^2=40000\)

b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^4-1\right)^2=18^8-18^8-1=-1\)

c) \(20^2+18^2+16^2+...+4^2+2^2-\left(19^2+17^2+...+3^2+1\right)\)

\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-...-3^2-1\)

\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+\left(16^2-15^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1\right)\)

\(=20+19+18+17+16+15+...+4+3+2+1\)

\(=\dfrac{\left(20+1\right).20}{2}=210\)

Chúc bạn học tốt!

Answer:

\(A=127^2+146.127+73^2\)

\(=127^2+2.127.73+73^2\)

\(=\left(127+73\right)^2\)

\(=200^2\)

\(=40000\)

\(B=9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=\left(9.2\right)^8-[\left(18^4\right)^2-1]\)

\(=18^8-18^8+1\)

\(=1\)

\(C=\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)

\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-15^2-...-3^2-1^2\)

\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(2^2-1^2\right)\)

\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(2-1\right)+\left(2+1\right)\)

\(=1.39+1.35+...+1.3\)

\(=39+35+...+3\)

Số số hạng \(\frac{39-3}{4}+1=10\) số hạng

Tổng \(\frac{\left(39+3\right).10}{2}=210\)

Ghi đầy đủ nha

bn có thể ghi rõ ràng đc ko?

a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\) 

=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\) 

=\(0+\dfrac{5}{13}\) 

=\(\dfrac{5}{13}\)

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

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17 +18 + 19 + 20 + 21 + 22 + 23 + 24 + 25=325

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17 +18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 = 325

Các bạn ơi cho mình vài **** đi mình ko có **** nào cả huhuhuhuhu...............................huhu