CMR:$1\frac{1}{2^3} 1\frac{1}{3^3} 1\frac{1}{4^3} ... 1\frac{1}{2019^3}< 2018\frac{1}{4}$AI NHANH TICK CHO 3TICK LUN

1

HM

Hazy Moon Hatsune Miku

4 tháng 4 2019

help me

DP

Đõ Phương Thảo

9 tháng 5 2019

CMR: D=$\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+.....+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}< \frac{1}{2}$

1

AH

Akai Haruma

Giáo viên

23 tháng 2 2020

Lời giải:

$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$

$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

Trừ theo vế:

$3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}$

$\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}$

Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$

$=4-\frac{6061}{4^{2019}}< 4$

$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)

NT

nhu thong Nguyen

4 tháng 5 2018

1) CMR:

$\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}$Ai nhanh mk tick

2) So sánh 2 số:

$A=\frac{5^{2018}-2016}{5^{2018}-2017}$$Với$$B=\frac{5^{2018}-2018}{5^{2018}-2019}$

5

DL

Duc Loi

4 tháng 5 2018

1) Đặt dãy trên là $A$

Theo bài ra ta có :

$A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}$

$\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$

$\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)$

2) $A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}$( 1 )

$B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}$( 2 )

Từ ( 1 ) và ( 2 ) $\Rightarrow$$A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B$

$\Rightarrow A< B$

Vậy $A< B.$

PG

Phạm Gia Khánh

4 tháng 5 2018

1) Ta có B =

$\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}$ < $\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$= $\frac{99}{100}$

=> B < 1 ( chứ không phải $\frac{1}{2}$ bạn nhé)

Sai thì thôi chứ mk chỉ làm rờ thôi

YP

Yến Phạm

12 tháng 3 2017

Cho À = $\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}$

CMR: A > $\frac{4}{3}$

Ai làm đúng vầ nhanh mk cho 10 tick lun

3

NM

Nguyễn Mạnh Khôi

12 tháng 3 2017

yêu cầu của đề bài là gì vậy bạn

NT

Nguyễn Thị Thu Thanh

12 tháng 3 2017

A = $\left(\frac{1}{11}+\frac{1}{12}+.........+\frac{1}{20}\right)$ + $\left(\frac{1}{21}+\frac{1}{22}+..........+\frac{1}{30}\right)$+ $\left(\frac{1}{31}+.....+\frac{1}{60}\right)$+ ... + $\frac{1}{70}$

Nhận xét:

$\frac{1}{11}$+ $\frac{1}{12}$+ ........ + $\frac{1}{20}$> $\frac{1}{20}$+$\frac{1}{20}$+........+$\frac{1}{20}$> $\frac{10}{20}$>$\frac{1}{2}$

$\frac{1}{21}+\frac{1}{22}+.......+\frac{1}{30}>\frac{30}{60}>\frac{1}{2}$

$\frac{1}{31}+......+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+.......+\frac{1}{60}>\frac{30}{60}>\frac{1}{2}$

A > $\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+......+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}>\frac{4}{3}$

NK

Nguyễn Kim Long

14 tháng 4 2019

$1.$Chứng minh rằng : $D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+.....+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}< \frac{1}{2}$

1

2T

๖²⁴ʱƘ-ƔℌŤ༉

14 tháng 8 2019

$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$

$\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

$\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

$-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}$

$\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}$

Đặt $M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}$

$\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}$

$\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}$

$-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}$

$\Rightarrow3M=1-\frac{1}{4^{2018}}$

$\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}$

$\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}$

$\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}$

$\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)$

V

☘️๖ۣۜвồ❤ċôηɠ❤αηɦ♔➻❥

11 tháng 12 2019

Tính P = $\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2019}}{\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+....\frac{1}{2018}}$

0

NQ

Nguyễn Quốc Hào

29 tháng 3 2020

Rút gọn:

$\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}$

1

HQ

Huỳnh Quang Sang

29 tháng 3 2020

Đặt $A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}$

$A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}$

$A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}$

$A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021$

HC

H cc

5 tháng 9 2016

1) $CMR:\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+......+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}$

Chỉ tick cho ai nhanh nhất

2

NT

Ngô Tấn Đạt

5 tháng 9 2016

Ta biến đổi vế phải :

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\\ $$\\ =\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}$

Vậy $\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}$

IM

Isolde Moria

5 tháng 9 2016

Ta có

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}$

$=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)$

$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}$

=> $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}$$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}$ ( đpcm )

F

❤Firei_Star❤

6 tháng 8 2018

Tìm x biết

a) $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}$

b) $\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}$

5

F

❤Firei_Star❤

7 tháng 8 2018

help me

PM

Phùng Minh Quân

7 tháng 8 2018

$a)$ Ta có :

$VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}$

$VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)$

$VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}$

$VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)$

Lại có :

$VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x$

$\Rightarrow$$x=2019$

Vậy $x=2019$

Chúc bạn học tốt ~

Tìm x biết:$(5^x+5^{x+1}+5^{x+2}):31=(3^{2x}+3^{2x+1}+3^{2x+2}):13$ CMR: $\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}+\frac{1}{3^{2019}}-\frac{1}{2}$ là một số âm Với giá trị nào của x thì biểu thức:$M=\frac{2|2018x-2019|+2019}{|2018x-2019|+1}$ đạt giá trị lớn nhất Cho a+b+c=2019...

AP

An Phương Hà

18 tháng 1 2019

Đọc tiếp

6

ZC

zZz Cool Kid_new zZz

18 tháng 1 2019

$\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}$

$=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}$

$=2+\frac{2017}{\left|2018x-2019\right|+1}$có giá trị lớn nhất

$\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}$có giá trị lớn nhất

$\Rightarrow\left|2018x-2019\right|+1$có giá trị nhỏ nhất

Mà $\left|2018x-2019\right|\ge0$

$\Rightarrow\left|2018x-2019\right|+1\ge1$

Dấu "=" xảy ra khi và chỉ khi:

$\left|2018x-2019\right|=0$

$\Leftrightarrow x=\frac{2019}{2018}$

Vậy $M_{MAX}=2019$tại $x=\frac{2019}{2018}$

ZC

zZz Cool Kid_new zZz

18 tháng 1 2019

$\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}$

$\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}$

$\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}$

$\Rightarrow5^x=3^{2x}$

Mà $\left(5;3\right)=1$

$\Rightarrow x=2x=0$