Tính: -1^2+2^2-3^2+4^2-...-17^2+18^2-19^2+20^2
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\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)
\(=39+35+....+7+3\)
\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)
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
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 \(\frac{\left(39+3\right).10}{2}=210\)
\(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\) = \(\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+1\)
= \(\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{20}\)
=\(20.\left(\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}+\frac{1}{20}\right)\)
=\(20.\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}\right)\)
Vì tử số gấp 20 lần mẫu số nên phân số này bằng 20