Question

giải giúp mình với 

\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)=?

Answer 1

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{2}.\frac{8}{9}\)

\(=\frac{4}{9}\)

Answer 2

Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9

2A=2/1.3+2/3.5+2/5.7+2/7.9

2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2A=1-1/9

2A=8/9

A=4/9

Answer 3

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}=\frac{4}{9}\)

k mik nha , mik k lại !!!!!!!!!!!!!!!!!!!!!!!

Answer 4

A=\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)

2xA=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+.....+\(\frac{2}{7.9}\)

2xA=1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{7}\)-\(\frac{1}{9}\)

2xA=1-\(\frac{1}{9}\)

    A=\(\frac{8}{9}\): 2

    A=\(\frac{4}{9}\) ỦNG HỘ MK NHA!!!!

Answer 5

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

= 1 - \(\frac{1}{9}\)

= \(\frac{8}{9}\)