1/ 1 + (-2) + 3 + (-4) + . . . + 17 + (-18)

= [1 + (-2)] + [3 + (-4)] + . . . + [17 + (-18)]

= (-1) + (-1) + . . . + (-1)

18 : 2 = 9

= (-1) . 9

= -9

2/ -1 + 2 - 3 + 4 + . . . - 99 + 100

= (-1 + 2) + (-3 + 4) + . . . (-99 + 100)

= 1 + 1 + . . . + 1

100 : 2 = 50

= 1 . 50

= 50

3/ 2 – 4 + 6 – 8 + . . . + 48 – 50 + 51 - 52

= (2 – 4) + (6 – 8) + . . . + (46 - 48) + (– 50 + 51 - 52)

= (-2) + (-2) + . . . + (-2) + (-51)

(48 - 2) : 2 + 1 = 24 : 2 = 12

= (-2) . 12 + (-51)

= (-24) + (-51)

= -75

4/ – 1 + 3 – 5 + 7 - . . . . + 97 – 99 + 101

= (–1 + 3) + (–5 + 7) + . . . . + (-93 + 95) + (97 – 99 + 101)

= 2 + 2 + . . . . + 2 + 99

(95 - 1) : 2 + 1 = 48 : 2 = 24

= 2 . 24 + 99

= 48 + 99

= 147

5/ 1 + 2 – 3 – 4 + . . . . + 97 + 98

= (1 + 2 – 3 – 4) + . . . . + (93 + 94 - 95 - 96) + (97 + 98)

= (-4) + . . . . + (-4) + 195

96 : 4 = 24

= (-4) . 24 + 195

= -96 + 195

= 99

1/ 1 + (-2) + 3 + (-4) + . . . + 17 + (-18)

=(1+3+5+..+17)+[-2+(-3)+(-4)+..+(-18)

goi ben tong so duong la A va tong so am la B ta co

A co ssh la :(17-1):2+1=9

B co ssh la :[-18-(-2)]:2+1=9

=>A=(17+1).9:2=81

=>B=[-18+(-2)].9:2=-72

=> A+B= 81+(-72)=9

vay tong bang 9

dãy trên có tất cả :(100-51):1+1=50 phân số

Ta có : 1/2:50=1/100

=>1/2=1/100+1/100+1/100+...+1/100(có tất cả 50 phân số 1/100)

Các phân số trong dãy S đều lớn hơn 1/100 ngoại trừ phân số cuối

=>dãy S >1/2

cac phan so 1/51;1/52;1/53;....1/99 đều lớn hơn 1/100. vậy S>1/100+1/100+....+1/100(co 50 phan so)=>S>50/100=1/2

Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)

\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)

=1