A=\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)+...+\(\frac{1}{99.100}\)

⇒A=\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)+...+\(\frac{1}{99}\)-\(\frac{1}{100}\)

⇒A=\(\frac{1}{10}\)-\(\frac{1}{100}\)

⇒A=\(\frac{9}{100}\)

Vậy A=\(\frac{9}{100}\)

B=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{97.99}\)

=\(\frac{1}{2}\).\((1-\frac{1}{3})\)+\(\frac{1}{2}.(\frac{1}{3}-\frac{1}{5})\)+...+\(\frac{1}{2}.(\frac{1}{97}-\frac{1}{99})\)

=\(\frac{1}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99})\)

=\(\frac{1}{2}.\frac{98}{99}\)

=\(\frac{49}{99}\)

Vậy B=\(\frac{49}{99}\)

a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99

A = 1 - 2 - 3 - 4 + 5 - 6 -  7 - 8 + ........... + 97 - 98 - 99 - 100 (100 số )

A = (1 - 2 - 3 - 4) + (5 - 6 - 7 - 8) + ................ + (97 - 98 - 99 - 100)

(25 cặp , tính bằng cách lấy số cả dãy chia cho số số của mỗi cặp )

A = (-8) . 25 

A = -200

59 nha

A = 1-2 + 3-4 + ......+ 99-100

= ( -1) + ( -1) +....+(-1) 

= (-1) x 50 

= -50

vì có tấ cả 50 số -1

a) A= 1-2+3-4+5-6+...+99-100

A = ( 1 - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ... + ( 99 - 100 )               ( có 50 cặp )

A = ( - 1 ) + ( -1 ) + ( -1 ) + ,.. + ( -1 )

A = ( - 1 ) . 50

A = -50

b) B = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + ... + 97 + 98 - 99 - 100

B = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ( 9 + 10 - 11 - 12 ) + ... + ( 97 + 98 - 99 - 100 )            ( có 25 cặp )

B = ( - 4 ) + ( - 4 ) + ( - 4 ) + ... + ( - 4 )

B = ( - 4 ) x 25 

B = -100