Ta có S = \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)⇒ S < \(\dfrac{1}{30}\cdot10+\dfrac{1}{40}\cdot10+\dfrac{1}{50}\cdot10=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

Vậy S < \(\dfrac{4}{5}\)

S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)

=>S>1/40*10+1/50*10+1/60*10=3/5

S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)

=>S<1/30*10+1/40*10+1/50*10=4/5

=>3/5<S<4/5

\(S=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)

ta có: \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\end{matrix}\right.\)

\(\Rightarrow S< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\Leftrightarrow5S< 4^{\left(1\right)}\)

Lại có: \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow S>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\Leftrightarrow5S>3^{\left(2\right)}\)

từ (1) và (2) => 3<5S<4

\(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

...

\(\dfrac{1}{40}=\dfrac{1}{40}\)

=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{41}>\dfrac{1}{50}\)

\(\dfrac{1}{42}>\dfrac{1}{50}\)

...

\(\dfrac{1}{50}=\dfrac{1}{50}\)

=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)

\(\dfrac{1}{51}>\dfrac{1}{60}\)

\(\dfrac{1}{52}>\dfrac{1}{60}\)

...

\(\dfrac{1}{60}=\dfrac{1}{60}\)

=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{10}{60}=\dfrac{1}{6}\)

=>\(S>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{3}{5}\)

\(\dfrac{1}{31}< \dfrac{1}{30}\)

\(\dfrac{1}{32}< \dfrac{1}{30}\)

...

\(\dfrac{1}{40}< \dfrac{1}{30}\)

=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

\(\dfrac{1}{41}< \dfrac{1}{40}\)

\(\dfrac{1}{42}< \dfrac{1}{40}\)

...

\(\dfrac{1}{50}< \dfrac{1}{40}\)

=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{51}< \dfrac{1}{50}\)

\(\dfrac{1}{52}< \dfrac{1}{50}\)

...

\(\dfrac{1}{60}< \dfrac{1}{50}\)

=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)

=>\(S< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{4}{5}\)

=>\(\dfrac{3}{5}< S< \dfrac{4}{5}\)

1/31>1/40

1/32>1/40

...

1/40=1/40

=>1/31+1/32+...+1/40>1/40*10=1/4

1/41>1/50

1/42>1/50

...

1/50=1/50

=>1/41+1/42+...+1/50>10/50=1/5

1/51>1/60

1/52>1/60

...

1/60=1/60

=>1/51+1/52+...+1/60>10/60=1/6

=>S>1/4+1/5+1/6=3/5

1/31<1/30

1/32<1/30

...

1/40<1/30

=>1/31+1/32+...+1/40<1/30*10=1/3

1/41<1/40

1/42<1/40

...

1/50<1/40

=>1/41+1/42+...+1/50<10/40=1/4

1/51<1/50

1/52<1/50

...

1/60<1/50

=>1/51+1/52+...+1/60<10/50=1/5

=>S<1/3+1/4+1/5=4/5

Ta có:\(\dfrac{31}{2}\).\(\dfrac{32}{2}\).\(\dfrac{33}{2}\).....\(\dfrac{60}{2}\)

=\(\dfrac{31.32.33.....60}{2^{30}}\)

=\(\dfrac{\left(1.2.3.....30\right).\left(31.32.33.....60\right)}{\left(1.2.3.....30\right).2^{30}}\)

=\(\dfrac{1.2.3.....60}{2.4.6.....60}\)

=\(\dfrac{\left(1.3.5.....59\right).\left(2.4.6.....60\right)}{2.4.6.....60}\)

=1.3.5.....59

Vậy (đpcm)

Đặt: \(\left\{{}\begin{matrix}A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\end{matrix}\right.\)

Ta có:

\(B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{59}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{30}\right)\)

\(=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\)

\(\Rightarrow B=A\)

Vậy \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\) (Đpcm)

Ta có:

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+......+\dfrac{1}{59.60}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{59}-\dfrac{1}{60}\)

= \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{59}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

- \(2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{60}\right)\) - \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)+ \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

= \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

Vậy\(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{59.60}\)