Cho: A=\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+........+\(\dfrac{1}{200}\). Chứng minh: \(\dfrac{1}{2}\)<A<1
Chứng minh : \(\dfrac{1}{2}< \dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+......................+\dfrac{1}{198}+\dfrac{1}{199}+\dfrac{1}{200}< \dfrac{100}{101}\)
Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Lại có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)
Vậy ...
Những dãy trên đều có 100 số hạng.
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 :
a) \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}=\dfrac{1}{101}\)+ \(\dfrac{1}{102}+...+\dfrac{1}{200}\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
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:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}+\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
⇒ \(ĐPCM\)
BT5 Chứng minh:
1) \(\dfrac{1}{2}< \dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{999}+\dfrac{1}{200}< 1\)
\(Tacó\)
\(\dfrac{1}{101}>\dfrac{1}{200}\)
\(\dfrac{1}{102}>\dfrac{1}{200}\)
...
\(\dfrac{1}{999}>\dfrac{1}{200}\)
Do đó :\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{999}+\dfrac{1}{200}>\dfrac{1}{200}+...+\dfrac{1}{200}=100.\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Ta lại có:
\(\dfrac{1}{102}< \dfrac{1}{101}\)
\(\dfrac{1}{103}< \dfrac{1}{101}\)
...
\(\dfrac{1}{200}< \dfrac{1}{101}\)
Do đó : \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{1}{101}.100=\dfrac{100}{101}< 1\)Vậy ...( theo tớ , cậu nên đánh dấu (1) và (2) rồi suy ra ) .. khẳng định trên , học tốt
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 = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)
\(\overline{50\text{ hạng tử }}\) \(\overline{50\text{ hạng tử }}\)
\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\)
\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Rightarrow P< \dfrac{5}{6}< 1\)
1.Chứng minh rằng :\(\dfrac{5}{8}< \dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{3}{4}\)
+)Đặt \(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)
\(A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...+\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+...+\dfrac{1}{200}\right)\)\(A>\dfrac{1}{125}.25+\dfrac{1}{150}.25+\dfrac{1}{175}.25+\dfrac{1}{200}.25=\dfrac{533}{840}>\dfrac{5}{8}\)
+)\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)
\(A=\left(\dfrac{1}{101}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...+\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...+\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...+\dfrac{1}{200}\right)\)\(A< \dfrac{1}{100}.20+\dfrac{1}{120}.20+\dfrac{1}{140}.20+\dfrac{1}{160}.20+\dfrac{1}{180}.20=\dfrac{1879}{2520}< \dfrac{3}{4}\)
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
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 \(⋮5=5\)
\(B=\left(\dfrac{1}{26}+...+\dfrac{1}{30}\right)+..+\left(\dfrac{1}{46}+..+\dfrac{1}{50}\right)\)
\(B>C=\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{16}{60}>\dfrac{16}{64}>\dfrac{2}{8}\\\dfrac{1}{7}+\dfrac{1}{9}=\dfrac{16}{63}>\dfrac{16}{64}>\dfrac{2}{8}\end{matrix}\right.\) \(\Rightarrow C>\dfrac{2}{8}+\dfrac{1}{8}+\dfrac{2}{8}=\dfrac{5}{8}\)
\(A>B>C>\dfrac{5}{8}\Rightarrow A>\dfrac{5}{8}\Rightarrow dpcm\Leftrightarrow dccm\)
Chứng minh rằng:
\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{7}{12}\)
Giúp mk nhé!
Ta có:
\(\dfrac{1}{101}>\dfrac{1}{150}\)
\(\dfrac{1}{102}>\dfrac{1}{150}\)
....
\(\dfrac{1}{150}=\dfrac{1}{150}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 số)=\(\dfrac{1}{3}\)
Ta có:
\(\dfrac{1}{152}>\dfrac{1}{200}\)
\(\dfrac{1}{153}>\dfrac{1}{200}\)
....
\(\dfrac{1}{200}=\dfrac{1}{200}\)
=>\(\dfrac{1}{151}+\dfrac{1}{153}+...+\dfrac{1}{120}>\dfrac{1}{120}+\dfrac{1}{120}+...+\dfrac{1}{120}\)(50 số)=\(\dfrac{1}{4}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}\)
=> \(A>\dfrac{7}{12}\)