giá trị biểu thức A=1/1+2+1/1+2+3+...+1/1+2+3+..+99 +1/50 là ?
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DQ
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
A
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
QT
1
MH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
DM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
LA
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
BT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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
M
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
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