Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(A=\frac{1}{2}-\frac{1}{17}\)

\(A=\frac{15}{34}\)

= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)

Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)

\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)

Vậy:..........................................(đpcm)

\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)

=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

= ​\(3\left(\frac{1}{2}-\frac{1}{17}\right)\)​​

=\(\frac{45}{34}\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)​

=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)

=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)

=3(1/2-1/17)

=45/34

cô Nhung ơi k đúng cho con đi cô pls

xét vế trái

ta có:đề\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\) 

\(=\frac{1}{2}-\frac{1}{17}< < \frac{1}{2}\)

vậy vế trái bé hơn \(\frac{1}{2}\)

P/S:  \(< < \)là luôn luôn bé hơn nha

k mình nha bạn

Thiengl2015#

Ta có :

\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}\)

Mà \(\frac{1}{2}-\frac{1}{17}< \frac{1}{2}\)

Nên \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}< \frac{1}{2}\left(đpcm\right)\)

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{14\cdot17}\)

=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

=\(\frac{1}{2}-\frac{1}{17}\)

=\(\frac{17}{34}-\frac{2}{34}\)

=\(\frac{15}{34}\)

Mà \(\frac{1}{2}=\frac{17}{34}>\frac{15}{34}\)

\(\Rightarrowđpcm\)

\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)

\(=3\cdot\frac{23}{200}\)

đúng

Đặt 3 ra ngoài

\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)

Vậy \(B=\frac{9}{25}\)

Chúc bn học tốt..!

a) = 2(1-1/2+1/2-1/3+...+1/19-1/20)

    = 2(1-1/20)

    = 2.19/20

    = 19/10

b) = 7(1/2-1/3+1/3-1/4+...+1/6-1/7)

   = 7(1/2 - 1/7)

   = 7.5/14

   = 5/2

c) = 1/2-1/5+1/5-1/8+...+1/14-1/17

   = 1/2 - 1/17

   = 15/34

Chúc bạn học tốt nhé

a)2/1.2+2/2.3+....+2/19.20

=2(1/1.2+1/2.3+....+1/19.20)

=2(1-1/2+1/2-1/3+.....-1/20)

=2(1-1/20)

2(19/20)=38/20=19/10

b)7/2.3+7/3.4+7/4.5+7/5.6+7/6.7

7(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)

7(1/2-1/3+1/3-1/4+.....-1/7)

7(1/2-1/7)

7(7/14-2/14)=7.5/14=35/14=5/2

c)3/2.5+3/5.8+3/8.11+3/11.14+3/14.17

1/2-1/5+1/5-1/8+......+1/14-1/17

1/2-1/17=17/34-2/34=15/34

Đây là toán lp 6 chứ bn

A = 2/1.2 + 2/2.3 + ...+2/19.20

Áp dụng công thức : n/n.(n+1)=(1/n)-(1/n+1)

Ta có : A= 1/1.2 + 1/2.3+...+1/19.20

           A= 1/1 - 1/2 + 1/2 - 1/3 + ...+ 1/19 - 1/20

           A= 1/1 - 1/20 

           A= 19/20

k cho mk nha bn

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

=\(\frac{3}{20}=0,15\)

\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)