Chứng minh rằng : \(\frac{1}{201}< \frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+\frac{1}{1005}< \frac{1}{201}\)Ai giải nhanh mình tick nha
CHỨNG MINH \(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{1500}>\frac{1}{3}\)
giúp mình với 1h mình hc r,cảm ơn nhaaaaaa
Ta có: 1/1500 = 1/1500
1/1001 > 1/1500
1/1002 > 1/1500
1/1003 > 1/1500 => 1/1001 + 1/1002 + 1/1003 + ... + 1/1499
. . . . . . . . . . . > 1/1500 + 1/1500 + 1/1500 + ... + 1/1500 (499 số hạng 1/1500)
1/1499 > 1/1500 > 499/1500
=> 1/1001 + 1/1002 + 1/1003 + ... + 1/1500 > 499/1500 + 1/1500 = 500/1500 = 1/3
Vậy 1/1001 + 1/1002 + 1/1003 + ... + 1/1500 > 1/3
k cho mình nha! Cảm ơn!
bạn có thể thêm dấu ngoặc vào sau chỗ:
1/1001 > 1/1500
1/1002 > 1/1500
1/1003 > 1/1500
. . . . . . . . . . . . .
1/1499 > 1/1500
Cho S = \(\frac{-1}{1001}+\frac{-1}{1002}+\frac{-1}{1003}+...+\frac{-1}{2000}\)
Chứng tỏ rằng S<\(\frac{-7}{12}\)
Chứng minh rằng:
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}<1\)
b) \(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}>\frac{13}{21}\)
Bạn đổi phân số thành / rồi tìm trên Google có đầy bài này rồi.
a, VT < 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/2007.2008
= 1-1/2+1/2-1/3+1/3-1/4+....+1/2007-1/2008 = 1-1/2008 < 1
=> ĐPCM
a) Ta có :1/22 + 1/32 + 1/42 + ... + 1/20082 < 1-1/2+1/2-1/3+...+1/2007-1/2008=1-1/2008<1
=> ĐPCM
Chứng minh rằng :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}_{ }\)
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)
\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)
Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006
=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)
=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)
=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)
=>A=1/1004+1/1005+.....+1/2006
Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )
Chứng minh rằng: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2002}\)
ta chuyển đề bài vế trái thành:
(1+1/2+1/3+1/4+...+1/2001+1/2002) - 2(1/2+1/4+1/6+...+1/2002)
=(1+1/2+1/3+....+1/2002) - (1+1/2+1/3+1/4+...+1/1001)
=1/1002+1/1003+...+1/2002
=> điều phải chứng minh
Chứng minh rằng \(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}<\frac{3}{4}\)
Tìm x,biết:
a)\(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
b)\(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
c)\(|x|-\frac{15}{2}=\frac{15}{4}\)
d)\(|\frac{3}{4}-x|+1|=\frac{3}{2}\)
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
mình làm luộn 2 câu còn lại nhé ^^
.c,\(|x|-\frac{15}{2}=\frac{15}{4}\)
.\(< =>|x|=\frac{15}{4}+\frac{15}{2}=\frac{15}{4}+\frac{30}{4}=\frac{45}{4}\)
.\(< =>\orbr{\begin{cases}x=\frac{45}{4}\\x=-\frac{45}{4}\end{cases}}\)
.d,\(|\frac{3}{4}-x|+1=\frac{3}{2}\)
.\(< =>|\frac{3}{4}-x|=\frac{3}{2}-1=\frac{3}{2}-\frac{2}{2}=\frac{1}{2}\)
.\(< =>\orbr{\begin{cases}\frac{3}{4}-x=\frac{1}{2}\\\frac{3}{4}-x=-\frac{1}{2}\end{cases}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\\x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\end{cases}}}\)
Chứng tỏ:
\(\frac{1}{1001}\)+\(\frac{1}{1002}\)+\(\frac{1}{1003}\)+...+\(\frac{1}{2000}\)>\(\frac{13}{21}\)
chứng minh rằng \(\frac{A}{B}\) là số nguyên
A = \(\frac{1}{1\times2}+\frac{1}{3\times4}+...+\frac{1}{2005\times2006}\)
B = \(\frac{1}{1004\times2006}+\frac{1}{1005\times2005}+...+\frac{1}{2006\times1004}\)