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: M= \(\dfrac{101^{102}+1}{101^{103}+1}\) và N= \(\dfrac{101^{103+1}}{101^{104}+1}\)
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.
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 < \(\dfrac{5}{6}\)
a: A>1/150*50+1/200*50=1/3+1/4=7/12
b: A>7/12
7/12>5/8
=>A>5/8
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: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)
\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)
. 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
Cho A= \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) Chứng tỏ A<1
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:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}+\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
⇒ \(ĐPCM\)
2. Chứng minh
a, \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\) < 1
b, \(\dfrac{1}{3}\)< \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{150}\)< \(\dfrac{1}{2}\)
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
2b.
Gọi tổng trên là $T$
Chứng minh vế đầu tiên:
Ta có:
$\frac{1}{101}> \frac{1}{150}$
$\frac{1}{102}> \frac{1}{150}$
....
$\frac{1}{149}> \frac{1}{150}$
$\Rightarrow T> \underbrace{\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}}_{50}=\frac{50}{150}=\frac{1}{3}$ (đpcm)
Chứng minh vế số 2:
$\frac{1}{101}< \frac{1}{100}$
$\frac{1}{102}< \frac{1}{100}$
....
$\frac{1}{150}< \frac{1}{100}$
$\Rightarrow T< \underbrace{\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}$ (đpcm)
