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Đặng Quốc Thắng
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A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

AN
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A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

Do Minh Duc
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A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

Nguyễn Gia Huy
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A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

phạm bảo duy
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Bùi Trần Khánh Huyền
Xem chi tiết

A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

Lan Anh Nguyễn
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Lan Anh Nguyễn
12 tháng 1 2016 lúc 16:51

Sai r bn ơi, là 2 ms đúng

 

A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

Đào Ngọc Quỳnh Anh
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Trần Anh
5 tháng 1 2016 lúc 14:27

= 1

tick đi mink giải thích cho .  hihihihihihihihiihihiiiiiiiiiiiiiiii

A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1

Quang tèo
Xem chi tiết
vu quang anh
19 tháng 1 2016 lúc 7:42

1,tớ làm trong violympic đúng đây