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�=131+132+133+…+160

�<(130+130+…+130)+(140+140+…+140)+(150+150+…+150)

�<1030+1040+1050<4860=45;

�>(140+140+…+140)+(150+150+…+150)+(160+160+…+160)

�>1040+1050+1060>35.

16 tháng 4 2023

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 < \(\dfrac{4}{5}\)

19 tháng 5 2021

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

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

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

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

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

...

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

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

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

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

...

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

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

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

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

...

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

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

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

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

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

...

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

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

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

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

...

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

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

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

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

...

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

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

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

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

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

25 tháng 3 2017

Giải:

Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

Ta có:

\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

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

Nhận xét:

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

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

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

\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)

Lại có:

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

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

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

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)

\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)

Từ \(\left(1\right)\)\(\left(2\right)\)

\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)

Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)

22 tháng 4 2018

Đặt A=131+132+133+...+159+160A=131+132+133+...+159+160

Ta có:

A=131+132+133+...+159+160A=131+132+133+...+159+160

⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)

Nhận xét:

131+132+...+140<130+130+...+130=13131+132+...+140<130+130+...+130=13

141+142+...+150<140+140+...+140=14141+142+...+150<140+140+...+140=14

151+152+...+160<150+150+...+150=15151+152+...+160<150+150+...+150=15

⇒A<13+14+15=4760<4860=45⇒A<13+14+15=4760<4860=45

⇒A<45(1)⇒A<45(1)

Lại có:

131+132+...+140>140+140+...+140=14131+132+...+140>140+140+...+140=14

141+142+...+150>150+150+...+150=15141+142+...+150>150+150+...+150=15

151+152+...+160>160+160+...+160=16151+152+...+160>160+160+...+160=16

⇒A>14+15+16=3760>3660=35⇒A>14+15+16=3760>3660=35

⇒A>35(2)⇒A>35(2)

Từ (1)(1)(2)(2)

⇒35<A<45⇒35<A<45

Vậy 35<131+132+133+...+159+160<4535<131+132+133+...+159+160<45

11 tháng 4 2017

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)

11 tháng 4 2017

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

NV
2 tháng 3 2023

Do mọi số hạng của B đều lớn hơn 0 nên \(B>0\)

Lại có:

\(B=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\) (30 số hạng)

\(\Rightarrow B< \dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{31}+...+\dfrac{1}{31}\) (30 số hạng)

\(\Rightarrow B< \dfrac{1}{31}.30\)

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

Vậy \(0< B< 1\)

\(\Rightarrow B\) nằm giữa 2 số tự nhiên liên tiếp nên B không phải là số tự nhiên

8 tháng 3 2023

Cảm ơn bn nhìu nheeee

 

30 tháng 10 2023

S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²

⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³

⇒ 2S/3 = S - S/3

= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)

= 1/3 - 1/3²⁰²³

⇒ S = (1/3 - 1/3²⁰²³) : 2/3

= (1 - 1/3²⁰²²) : 2

Lại có: 1 - 1/3²⁰²² < 1

⇒ S < 1/2