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

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

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

Vì tổng A có 100 số hạng,2 số hạng tạo thành 1 cặp nên 100 số hạng tạo thành 50 cặp

=> A = ( -1 ) . 50

=> A = -50

 b) B = 1 + 3 - 5 - 7 + 9 + 11 - .... - 397 - 399

=> B = ( 1 + 3 - 5 - 7 ) + ( 9 + 11 - 13 - 15 ) + ... + ( 393 + 395 - 397 - 399 )

=> B = ( -8 ) + ( -8 ) + ... + ( -8 )

Vì tổng B có 200 số hạng,4 số hạng tạo thành 1 cặp nên 200 số hạng tạo thành 50 cặp

=> B = ( -8 ) . 50 

=> B = -400

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

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

=> C = 0 + 0 + ... + 0

=> C = 0

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

A = ( 1 -2 ) + ( 3 - 4 ) + ..... + ( 99 - 100 )  ( 50 nhóm )

A = 1 + 1 + .... + 1 ( 50 số 1 )

A = 1 . 50

A = 50

a)\(\frac{32}{64}-\frac{16}{64}+\frac{8}{64}-\frac{4}{64}+\frac{2}{64}-\frac{1}{64}\le\frac{1}{3}\)

\(\Rightarrow\frac{32-16+8-4+2-1}{64}=\frac{23}{64}\)\

\(\Rightarrow\frac{23}{64}=0,359375;\frac{1}{3}=0,33333...\)

đề sao lạ vậy

@ Bùi Long Vũ tinh sai roi kia:

32-16+8-4+2-1=21 mak 

biết mỗi câu a

ukm thế cx dc

cau a bang 101/200, k cho minh nha'

1/3 + 1/15 + 1/35 + 1/63 + 1/99 + 9999

= 1/3 + ( 1/5 + 1/35 + 1/63 ) + 1/99 = 9999

= 1/3 + 1111/9999 + 1/99

= 3333/9999 + 1111/9999 +101/9999

= 4545/9999

4546/9999

có thể giải hẳn ra được ko cái này tính máy tính cũng ra

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)

\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)

\(B=\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot...\cdot\left(1-\dfrac{1}{1326}\right)\\ =\dfrac{20}{21}\cdot\dfrac{27}{28}\cdot\dfrac{35}{36}\cdot...\cdot\dfrac{1325}{1326}\\=\dfrac{40}{42}\cdot\dfrac{54}{56}\cdot\dfrac{70}{72}\cdot...\cdot\dfrac{2650}{2652}\\ =\dfrac{5\cdot8}{6\cdot7}\cdot\dfrac{6\cdot9}{7\cdot8}\cdot\dfrac{7\cdot10}{8\cdot9}\cdot...\cdot\dfrac{50\cdot53}{51\cdot52}\\ =\dfrac{5\cdot8\cdot6\cdot9\cdot7\cdot10\cdot...\cdot50\cdot53}{6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot...\cdot51\cdot52}\\ =\dfrac{\left(5\cdot6\cdot7\cdot...\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot...\cdot53\right)}{\left(6\cdot7\cdot8\cdot...\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot...\cdot52\right)}=\dfrac{5\cdot53}{51\cdot7}=\dfrac{265}{357} \)