\(A\) = \(\dfrac{1}{10}\)+\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\) chứng tỏ A < \(\dfrac{51}{20}\)
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
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2022
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
27 tháng 2 2017
Đặt \(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)
Nhận xét:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+...+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)
\(\Rightarrow A>\frac{4}{3}\)
Vậy \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}>\frac{4}{3}\) (Đpcm)
NT
27 tháng 2 2017
\(A=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+....+\dfrac{1}{70}\\ =\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\right)+\left(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}\right)+\left(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}\right)+....+\dfrac{1}{70}\\ \)
\(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}=\dfrac{1}{2}\)
\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+....+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)
\(\Rightarrow A>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{61}+...+\dfrac{1}{70}>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{3}\)
Chúc bạn học tốt !!!!!!
NH
0
mk pít lm phần a nhưng k có thời gian
AM
8 tháng 6 2021
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
8 tháng 6 2021
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
LA
7 tháng 4 2017
\(\dfrac{1}{12}>\dfrac{1}{22};\dfrac{1}{13}>\dfrac{1}{22};...;\dfrac{1}{21}>\dfrac{1}{22};\dfrac{1}{22}=\dfrac{1}{22}\)
\(\Rightarrow\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{22}>\dfrac{1}{22}.11\) (do A có 11 số hạng)
\(\Leftrightarrow A>\dfrac{11}{22}=\dfrac{1}{2}\) ( đpcm)
AH
Akai Haruma
Giáo viên
11 tháng 8 2023
Lời giải:
$A=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....+\frac{69}{7^{70}}$
$7A=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$
$\Rightarrow 6A=7A-A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{69}}-\frac{69}{7^{70}}$
$42A=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$
$\Rightarrow 36A=42A-6A=1-\frac{69}{7^{69}}+\frac{69}{7^{70}}<1$
$\Rightarrow A< \frac{1}{36}$