Question

CMR :

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+..........+\frac{3}{99.101}\)     \(< 1,5\)

Answer 1

Ta có : 

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(=\)\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(=\)\(\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\)\(\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(=\)\(\frac{3}{2}.\frac{100}{101}< \frac{3}{2}.\frac{100}{100}=1,5\)

\(\Rightarrow\)\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}< 1,5\) ( đpcm ) 

Vậy \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}< 1,5\)

Chúc bạn học tốt ~ 

Answer 2

tham khảo ở đây nhé

https://olm.vn/hoi-dap/question/185679.html

^^

Answer 3

bn Nguyễn Hồng Hà My ms hok lớp 5 mak

mk cần cách giải đầy đủ ko cóp mạng mk cho 2 tk

Answer 4

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+........+\frac{3}{99.101}=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{99.101}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{101}\right)=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}< \frac{150}{100}=1.5\)(đpcm)

Answer 5

=\(3.\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...........+\frac{1}{99.101}\right]\)

= 3 . [ \(\frac{1}{2}.\)(\(1-\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\)\(\frac{1}{5}-\frac{1}{7}\)+ ............+\(\frac{1}{99}-\frac{1}{101}\)) ]

= 3 . [\(\frac{1}{2}.\) 1 -\(\frac{1}{2}.\)  \(\frac{1}{3}\)+\(\frac{1}{2}.\) \(\frac{1}{3}\) - \(\frac{1}{2}.\) \(\frac{1}{5}\)+ \(\frac{1}{2}.\) \(\frac{1}{5}\) - \(\frac{1}{2}.\) \(\frac{1}{7}\)+ .........+\(\frac{1}{2}.\) \(\frac{1}{99}\) - \(\frac{1}{2}.\) \(\frac{1}{101}\)]

= 3 . [ \(\frac{1}{2}.\) 1  - \(\frac{1}{2}.\)\(\frac{1}{101}\) ]

=3.[ \(\frac{1}{2}.\) ( 1 - \(\frac{1}{101}\) )

= 3 . [ \(\frac{1}{2}.\) \(\frac{100}{101}\)]

= 3 . \(\frac{50}{101}\)

= \(\frac{150}{101}\)

Ta có : 1,5 = \(\frac{15}{10}=\frac{150}{100}\)

Vì \(\frac{150}{101}\) < \(\frac{150}{100}\)

\(\Rightarrow\)\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+..........+\frac{3}{99.101}\)< 1,5

Vậy \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+..........+\frac{3}{99.101}\) < 1,5