Ta có :

\(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+...+\dfrac{1}{1985}\)

\(A=\dfrac{1}{5}+\dfrac{1}{3.5}+\dfrac{1}{5.5}+\dfrac{1}{7.5}+...+\dfrac{1}{397.5}\)

\(\Rightarrow5A=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)

\(5A-1=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)

\(5A-1=\dfrac{1}{3}+\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)+\left(\dfrac{1}{11}+\dfrac{1}{13}+...+\dfrac{1}{27}\right)+\)

\(\left(\dfrac{1}{29}+\dfrac{1}{31}+...+\dfrac{1}{81}\right)+\left(\dfrac{1}{83}+\dfrac{1}{85}+...+\dfrac{1}{243}\right)+...+\dfrac{1}{397}\)

\(\Rightarrow5A-1>\dfrac{1}{3}+\dfrac{1}{9}.3+\dfrac{1}{27}.9+\dfrac{1}{81}.27+\dfrac{1}{243}.81=\dfrac{1}{3}.5=\dfrac{5}{3}\)

\(\Rightarrow5A-1>\dfrac{5}{4}\Rightarrow5A>\dfrac{9}{4}\)

\(\Rightarrow A>\dfrac{9}{4}:5=\dfrac{9}{20}\Rightarrow\left(dpcm\right)\)

\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)

\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)

\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)

Do đó: A=B

Giải:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{5}+\dfrac{1}{13}\cdot3+\dfrac{1}{61}\cdot3\\ =\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

=> Điều phải chứng minh

\(\dfrac{5}{2x1}+\dfrac{4}{1x11}+\dfrac{3}{11x2}+\dfrac{1}{2x15}+\dfrac{13}{15x4}+\dfrac{15}{4x13}\)

=7x(\(\dfrac{5}{2x7}+\dfrac{4}{7x11}+\dfrac{3}{11x14}+\dfrac{1}{14x15}+\dfrac{13}{15x28}+\dfrac{15}{28x43}\))

=7x\(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{43}\)=7x(\(\dfrac{1}{2}-\dfrac{1}{43}\))

=7x\(\dfrac{41}{86}\)

=\(\dfrac{287}{86}\)

5/2x1+4/1x11+3/11x2+1/2x15+13/15x4+15/4x43=7x(5/2x7+4/7x11+3/11x14+1/14x15+13/15x28+15/28x43)=7x(1/2-1/7+1/7-1/11+1/11-1/14+1/14+1/15+1/15-1/28+1/28-1/43)=7x(1/2-1/43)=7x41/86=287/86

Ta có :

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét :

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

Chứng minh 1 bất đẳng thức cơ bản sau:\(\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}\)

Thật vậy: \(\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{n^2+n^2+2n+1}=\dfrac{1}{2n^2+2n+1}< \dfrac{1}{2n^2+2n}=\dfrac{1}{2n\left(n+1\right)}\)

Thay vào bài toán \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+\dfrac{1}{3^2+\left(3+1\right)^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)

\(< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}-\dfrac{1}{2\left(n+1\right)}< \dfrac{1}{2}\left(đpcm\right)\)

\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)

\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(M=2.\dfrac{3}{16}\)

\(M=\dfrac{3}{8}\)

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)