A = \(\dfrac{1}{101}\)+ \(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+ ... + \(\dfrac{1}{200}\). Chứng minh:
a) A > \(\dfrac{7}{12}\)
b) A > \(\dfrac{5}{8}\)
c) A < \(\dfrac{5}{6}\)
So sánh:
a)\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) với 1
b)\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{149}+\dfrac{1}{150}\) với\(\dfrac{1}{3}\)
c)\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) với \(\dfrac{7}{12}\)
c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)
Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)
Tương tự
\(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)
\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2)
Từ (1) và (2) ta được
\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)
\(\overline{50\text{ hạng tử }}\) \(\overline{50\text{ hạng tử }}\)
\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\)
\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Rightarrow P< \dfrac{5}{6}< 1\)
Cho A=\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
CMR: a) A>\(\dfrac{7}{12}\)
b) A>\(\dfrac{5}{8}\)
a. ta có
1/101 > 1/150
1/102> 1/150
...>1/150
1/150 = 1/150
=> 1/101 + 1/102 + .... + 1/150 > 1/150 +1/150+....+1/150(50 số hạng )= 1/3
ta có
1/151 >1/200
1/152 > 1/200
..>1/200
1/200 = 1/200
=> 1/151 + 1/152+....+1/200 > 1/200+1/200+ ...+1/200( 50 số hạng) = 1/4
==> 1/101 + 1/102+....+1/200 > 1/3 +1/4
==> A > 7/12
b, A =(1/101+1/102+....+1/150)+(1/151+1/152+.....+1/200)
A>1/150.50+1/200.50=1/3+1/4=7/12
b tách A thành bốn nhóm rồi cũng làm như trên,ta có
A>25/125+25/150+25/175+25/200=(1/5+1/6+1/7)+1/8
=107/210+1/8>1/2+1/8=5/8
Chứng minh rằng \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{7}{12}\)
Ta có:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\) (có 50 số hạng)
⇔ \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{3}\) \(\left(1\right)\)
\(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\) (có 50 số hạng)
⇔ \(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{4}\) \(\left(2\right)\)
Từ (1) và (2), cộng vế theo vế. Ta được:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}+\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
⇒ \(ĐPCM\)
Cho C=\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) Chứng minh rằng C > 5/8
Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)
\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)
\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)
\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)
hay \(C>\dfrac{5}{8}\)(đpcm)
Chứng minh rằng \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
A>\(\dfrac{5}{8}\)
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+....+\dfrac{1}{200}\)
\(A=\sum\limits^{200}_{x=101}\left(\dfrac{1}{x+1}\right)=0,6857275648\)
Có: \(\dfrac{5}{8}=0.625\)
mà \(0,685...>0,625\)
\(\Rightarrow A>\dfrac{5}{8}\)
p/s: đây chỉ là 1 cách thoy, có cần lm cách khác k?
Lời Giải
Hay sử lý các con số khi không cần máy tính
\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)
dãy A có 100 số hạng \(⋮4=25\)
\(A=\left(\dfrac{1}{101}+...+\dfrac{1}{104}\right)+\left(\dfrac{1}{105}+..+\dfrac{1}{108}\right)+..+\left(\dfrac{1}{197}+\dfrac{1}{200}\right)\) Bao gồm (..)
\(A>B=\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\right)\)
dãy A có 25 số hạng \(⋮5=5\)
\(B=\left(\dfrac{1}{26}+...+\dfrac{1}{30}\right)+..+\left(\dfrac{1}{46}+..+\dfrac{1}{50}\right)\)
\(B>C=\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{16}{60}>\dfrac{16}{64}>\dfrac{2}{8}\\\dfrac{1}{7}+\dfrac{1}{9}=\dfrac{16}{63}>\dfrac{16}{64}>\dfrac{2}{8}\end{matrix}\right.\) \(\Rightarrow C>\dfrac{2}{8}+\dfrac{1}{8}+\dfrac{2}{8}=\dfrac{5}{8}\)
\(A>B>C>\dfrac{5}{8}\Rightarrow A>\dfrac{5}{8}\Rightarrow dpcm\Leftrightarrow dccm\)
Cho A= \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) Chứng tỏ A<1
Cho S = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
So sánh: a, S và \(\dfrac{1}{2}\)
b, S và 1
cho A = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}\), e hãy chứng minh A>\(\dfrac{2}{3}\)
refer
https://hoc247.net/hoi-dap/toan-6/chung-minh-1-101-1-102-1-103-1-104-1-299-1-300-2-3-faq302038.html
thamkhao
https://hoc247.net/hoi-dap/toan-6/chung-minh-1-101-1-102-1-103-1-104-1-299-1-300-2-3-faq302038.html
1)
Cho \(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{200}\)
Chứng minh: \(A>\dfrac{9}{10}\)
2)
Cho \(B=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
Chứng minh: \(B>\dfrac{7}{12}\)
HELP ME!!!!!!!!
\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )
Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)
...
\(\dfrac{1}{199}>\dfrac{1}{200}\)
Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...