`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`

`=1/2xx2/3xx3/4xx4/5`

`=[1xx2xx3xx4]/[2xx3xx4xx5]`

`=1/5`

`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`

`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`

`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`

`=1/100`

A = 0-1 + 2-3 + 4-5 +...+ 2017-2018 

=> A = (-1) + (-1) + (-1)  +...+ (-1) (Có 1009 số hạng)

=> A = 1009.(-1)

=> A = -1009

B = 1-3+5-7+ 9-11+....+2005-2007 

=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)

=> B = 502.(-2)

=> B = -1004

C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102

=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)

=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)

=> C = (-9).17

=> C = -153

C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

  =\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\frac{99}{100}\)

  =\(\frac{-98}{100}=\frac{-49}{50}\)

C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1 
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1) 
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A 
Dễ thấy 1/2.1 = 1/1 - 1/2 
1/3.2 = 1/2 - 1/3 
..................... 
1/99.98 = 1/98 - 1/99 
1/100.99 = 1/99 - 1/100 
=> cộng từng vế với vế ta

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)

\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=0+0+0+0-\frac{1}{16}\)

\(=\frac{-1}{16}\)

Các anh các cj giúp em nhé 
Em cảm ơn trước

Giải:

a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\) 

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\) 

\(=\dfrac{1.2.3.4}{2.3.4.5}\) 

\(=\dfrac{1}{5}\) 

b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\) 

\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\) 

\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\) 

\(=\dfrac{1}{100}\) 

Chúc bạn học tốt!

A)

\(\left(1-\dfrac{1}{2}\right)x\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x\left(1-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}x\dfrac{1}{3}x\dfrac{1}{4}x\dfrac{1}{5}\)

\(=\dfrac{1x2x3x4}{2x3x4x5}\)

\(=\dfrac{1}{5}\)

B)

\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right)......\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)

\(\dfrac{1}{4}.\dfrac{1}{7}.....\dfrac{94}{97}.\dfrac{97}{100}\)

\(\dfrac{1}{100}\)

học tốt nhé!!!

a)\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)=\(7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)=\(\left(7\frac{4}{9}-3\frac{4}{9}\right)+4\frac{7}{11}\)= 4+\(4\frac{7}{11}\)=\(8\frac{7}{11}\)

b)\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)=\(\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5+\frac{7}{9}\)=\(\frac{-7}{9}.1+5+\frac{7}{9}\)=\(\frac{-7}{9}+\frac{7}{9}+5\)=\(\left(\frac{-7}{9}+\frac{7}{9}\right)+5\)= 0+5=5

c)\(50\%.1\frac{1}{3}.10\frac{7}{35}.0,75\)= \(\frac{1}{2}.\frac{4}{3}.10\frac{1}{5}.\frac{3}{4}\)=\(\frac{1}{2}.\frac{4}{3}.\frac{51}{5}.\frac{3}{4}\)=\(\frac{1.4.51.3}{2.3.5.4}\)=\(\frac{51}{2.5}\)=\(\frac{51}{10}\)

d)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)

a) \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=\left(\frac{67}{9}+\frac{51}{11}\right)-\frac{31}{9}\)

\(=\left(\frac{67}{9}-\frac{31}{9}\right)+\frac{51}{11}\)

\(=\frac{36}{9}+\frac{51}{11}\)

\(=\frac{95}{11}=8\frac{7}{11}\)

b) \(-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+5\frac{7}{9}\)

\(=-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+\frac{52}{9}\)

\(=-\frac{7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)

\(=-\frac{7}{9}.1+\frac{52}{9}\)

\(=-\frac{7}{9}+\frac{52}{9}\)

= 5

d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)

\(=1.\left(1-\frac{1}{43}\right)\)

\(=1.\frac{42}{43}\)

\(=\frac{42}{43}\)

Cảm ơn

a) B= 1 + 3 + 7 + 9 +...+ 99

= ( 1 + 99 ) + ( 97 + 3 ) + ( 95 + 5 ) + ( 93 + 7 ) + ( 91 + 9 ) + ( 89 + 11 ) + ... ,

(Tổng cộng có 25 cặp có tổng là 100.)
= 25 x 100.
B = 2500.

b,    C = 1+4+7+10+..............+100

Số số hạng là : 

(100-1) : 3 + 1 = 34 (số)

C = (100+1) x 34 : 2 

C = 101 x 34  : 2

C = 3434 : 2

C = 1717

a: Số số hạng là:

(99-1):2+1=50(số)

Tổng là:

\(\dfrac{100\cdot50}{2}=100\cdot25=2500\)

a. 1+3+5+....+99=2500

b 1+4+7+....+99+102=15555

Mình ko tiện giải chi tiết vì lâu

Thông cảm!