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
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 \(M=\dfrac{101^{102}+1}{101^{103}+1}\) và N = \(\dfrac{101^{103}+1}{101^{104}+1}\) So sánh M và N.
Tham khảo:
https://hoc247.net/hoi-dap/toan-6/so-sanh-m-101-102-1-101-103-1-va-n-101-103-1-101-104-1--faq225210.html
so sánh: M= \(\dfrac{101^{102}+1}{101^{103}+1}\) và N= \(\dfrac{101^{103+1}}{101^{104}+1}\)
So sánh A và B :
A = \(\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
B = \(\dfrac{1}{2^2.3.5^2.7}\)
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
1) Cho A= \(\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{101}}+\dfrac{1}{\sqrt{102}}+\dfrac{1}{\sqrt{103}}+\dfrac{1}{\sqrt{104}}\)
và B= \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{100^2}\right)\)
So sánh A và B.
So sánh : a) (-39)9 và (-18)13
b)\(\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\) và \(\dfrac{1}{2^2.3.5^2.7}\)
Cho A= \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) Chứng tỏ A<1
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
cmr \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\)
Ta có: \(\dfrac{1}{101}>\dfrac{1}{200}\)
Tương tự ta có: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)