S = 1/2.5 +1/5.8 +1/8.11+1/11.14+1/14.17+1/17.20

S=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)

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

S=1/3.(10/20-1/20)

S=1/3.9/20

S= 3/20

k nha

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

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

mk đầu tiên đó

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

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

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

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

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

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)

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

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

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

Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)

A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

sai ùi 

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

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

\(A=\frac{1}{3}.\frac{9}{20}\)

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

Vì \(\frac{3}{20}< 1\Rightarrow A< 1\)

Cau a) 1/1.2 +1/2.3 +1/3.4+...+1/99.100= 1/1-1/2+1/2-1/3+...+1/99-1/100

              =1/1-1/100=99/100

              99/100<1 thì 1/1.2 +1/2.3+1/3.4+...+1/99.100<1

Câu b): Ta có: 1/2^2<1/1.2

                      1/3^2<1/2.3

                       ...............(so sánh như vậy với các số khác)

                       1/2016^2<1/2015.2016

                       Áp dụng của câu a ta thêm vào sau về thành: 1/1.2+1/2.3+1/3.4+...+1/2015.2016

                       =1/1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016

                       =1/1-1/2016

                       =2015/2016<1

                       Ma :1/2^2+1/3^2+1/4^2+...+1/2016^2<1/1.1+1/2.3+1/3.4+...+1/2015.2016

                       Nen:1/1^2+1/3^2+1/4^2+...+1/2016^2<1

Câu c: 3/2.5+3/5.8+3/8.11+3/11.14+3/14.1

           =1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17

           =1/2-1/17=15/34<1/2

           Nen:3/2.5+3/5.8+3/8.11+3/11.14+3/14.17<1/2

           (khoảng cách từ 2-5;5-8;8-11;11-14 là 3 nên số 3 biến đổi thành số 1 ở dòng thứ nhất xuống dòng thứ 2)

x.\(\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)=-1\frac{3}{5}\)

x.\(\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\right)=\frac{-8}{5}\)

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

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

x.\(\left(\frac{17}{34}-\frac{2}{34}\right)=\frac{-8}{5}\)

x.\(\frac{15}{34}=\frac{-8}{5}\)

x\(=\frac{-8}{5}:\frac{15}{34}\)

x\(=\frac{-8}{5}.\frac{34}{15}\)

x\(=\frac{-272}{75}\)

Vậy x\(=\frac{-272}{75}\)

Ta có :

\(2\cdot x+\frac{x}{2}+\frac{x}{4}+1=100\)

\(\frac{11}{4}x=100-1=99\)

Số cò là:

\(99\div\frac{11}{4}=36\left(con\right)\)

ĐS:..........

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.