Question

[Nguyễn Thế Mạnh]

giải rõ cách giùm mình nha.

1/3x5 + 1/5x7 + 1/7x9 +...+ 1/97x99a

Answer 1

[Hoàng Ninh]

\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+..........+\frac{1}{97x99}\)

= \(1-\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-........-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

= \(1-\frac{1}{3}-\frac{1}{99}\)

= \(\frac{99}{99}-\frac{33}{99}-\frac{1}{99}\)

= \(\frac{65}{99}\)

Answer 2

[# APTX _ 4869 _ : ( $>$ )]

\(\frac{1}{3}\)*5+\(\frac{1}{5}\)*7+\(\frac{1}{7}\)*9*...*\(\frac{1}{97}\)*99

=\(\frac{5}{3}\)*\(\frac{7}{5}\)*\(\frac{9}{7}\)*...*\(\frac{99}{97}\)

=\(\frac{99}{3}\)

đúng thì nha