A = ( 1 + 1/3 ) + ( 1 + 1/15 ) + ( 1 + 1/35 ) + ( 1 + 1/63 ) + .... + ( 1 + 1/9999 )

A = ( 1 + 1 + 1 + ...) + ( 1/3 + 1/15 + 1/35 + 1/63 + ....+ 1/9999 )

tự làm tiếp

kết quả bằng 1 / 3 nhé bạn

9999 / 36 - 15*11 = 1111 / 4 - 165 = 1111 / 4 - 660 / 4 = 451 / 4 = 112 ,75

A = \(\frac{2}{3}+\frac{3}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)

A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)

A = \(1-\frac{1}{101}\)

A = \(\frac{100}{101}\)

hồ thu giang làm đúng rồi

Bài 6:

Đổi 4 giờ 30 phút = 4,5 giờ

Vận tốc xe đạp là:

54 : 4,5 = 12 (km/h)

Vận tốc xe máy là: 54 : 1,5 = 36 (km/h)

Mỗi giờ xe máy đi hơn xe đạp số ki-lô-mét là:

36 - 12 = 24 (km/h)

Đáp số: 24 km/h 

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Câu tl ko liên quan: Mk thấy đề cuối học kì nó khác khác mà bn ơi?

Dù có ở tỉnh nào thì cũng giúp đi

biết mỗi câu a

ukm thế cx dc

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

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

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

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