A                   =      \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)+......+\(\dfrac{1}{1024}\)

A \(\times\)2           = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+...+\(\dfrac{1}{512}\)

A\(\times\)2 - A     =   1 - \(\dfrac{1}{1024}\)

A               = \(\dfrac{1023}{1024}\)

B = \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+...+\(\dfrac{1}{98\times99}\)+\(\dfrac{1}{99\times100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{99}\)+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

B = 1 - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C= \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+...+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)

C = \(\dfrac{1}{2\times3}\) +\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+...+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)

C = \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)

C = \(\dfrac{1}{2}\) - \(\dfrac{1}{11}\)

C = \(\dfrac{9}{22}\)

1 found out

2 dress up

3 applied for

4 go on

5 cheered up

6 take off

7 pull down

8 set up

9 get over

10 turn back

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

a/ Ta có:

- ABCD là hình bình hành => \(AD=BC;AD\text{ // }BC\) 

- Xét △ADH và △BKC có:

 + \(AD=BC\left(cmt\right)\)

 + \(\hat{ADH}=\hat{CBK}\left(slt\right)\)

=> \(\text{△ADH = △CBK}\left(c.h-g.n\right)\)

Vậy: \(...