Bài 15.
a) So sánh \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
b) Cho biểu thức A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}.\)Chứng tỏ A < 1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TT
21 tháng 7 2020
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
gọi
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
VÌ \(\frac{2019}{2020}< 1\Rightarrow A< 1\)
VẬY \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)
21 tháng 7 2020
1. a) P = 4 - ( x - 2 )32
( x - 2 )32 ≥ 0 ∀ x => - ( x - 2 )32 ≤ 0 ∀ x
=> 4 - ( x - 2 )32 ≤ 4 ∀ x
Dấu bằng xảy ra <=> x - 2 = 0 => x = 2
Vậy PMax = 4 khi x = 2
b) Q = 20 - | 3 - x |
| 3 - x | ≥ 0 ∀ x => - | 3 - x | ≤ 0 ∀ x
=> 20 - | 3 - x | ≤ 20 ∀ x
Dấu bằng xảy ra <=> 3 - x = 0 => x = 3
Vậy QMax = 20 khi x = 3
c) C = \(\frac{5}{\left(x-3\right)^2+1}\)
Để C có GTLN => ( x - 3 )2 + 1 nhỏ nhất dương
=> ( x - 3 )2 + 1 = 1
=> ( x - 3 )2 = 0
=> x - 3 = 0
=> x = 3
=> CMax = \(\frac{5}{\left(3-3\right)^2+1}=\frac{5}{1}=5\)khi x = 3
14 tháng 5 2019
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
14 tháng 5 2019
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
S
12 tháng 5 2017
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
Q
12 tháng 5 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
4 tháng 2 2017
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
B
8 tháng 2 2020
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
---------------------
Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
3 tháng 1 2017
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
Bài 15 :
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)
b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)
Tới đây là so sánh đi nhé
Cái này mình làm hôm qua rồi mà '-'
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A< 1\)
b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)
\(2A-A=A\)
\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)
\(=1-\frac{1}{2^{1000}}\)
\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)