\(\dfrac{2002}{2003}\) và \(\dfrac{14}{13}\)
So sánh
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
HT
3
DE
8 tháng 8 2019
Ta có
\(\frac{2002}{2003}< 1< \frac{14}{13}\)
\(\frac{\Rightarrow2002}{2003}< \frac{14}{13}\)
8 tháng 8 2019
có\(\hept{\begin{cases}\frac{2002}{2003}< 1\\\frac{14}{13}>1\end{cases}\Leftrightarrow\frac{2002}{2003}< 1< \frac{14}{13}}\)
vậy \(\frac{2002}{2003}< \frac{14}{13}\)
HG
1
20 tháng 7 2016
So sánh:
Ta có:\(\hept{\begin{cases}\frac{2002}{2003}< \frac{2003}{2003}=1\\\frac{14}{13}>\frac{13}{13}=1\end{cases}}\)
\(\Rightarrow\frac{14}{13}>\frac{2002}{2003}\)
12 tháng 9 2017
a)Ta có :
\(-\dfrac{265}{317}< -\dfrac{83}{317}< -\dfrac{83}{111}\Rightarrow-\dfrac{265}{317}< -\dfrac{83}{111}\)
b)Ta có :
\(\dfrac{2002}{2003}< 1< \dfrac{14}{13}\Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)
c)Ta có :
\(\dfrac{-1}{-3}=\dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< 0< \dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< \dfrac{1}{3}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
DC
2
H
1 tháng 3 2023
ta có : `12/14 = 6/7`
`13/24=(13xx7)/(24xx7)= 91/168`
`6/7=(6xx24)/(7xx24)= 144/168`
mà : `91<144`
`=> 13/24 < 12/14`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 3 2023
\(\dfrac{13}{24}\) = \(\dfrac{13\times7}{24\times7}\) = \(\dfrac{91}{168}\)
\(\dfrac{12}{14}\) = \(\dfrac{12\times12}{14\times12}\) = \(\dfrac{144}{168}\)
\(\dfrac{91}{689}< \dfrac{144}{168}\)
\(\dfrac{13}{24}\) < \(\dfrac{12}{14}\)
25 tháng 8 2017
a)\(\dfrac{2002}{2003}\) và \(\dfrac{14}{13}\)
\(\dfrac{2002}{2003}< 1;\dfrac{14}{13}>1\)
\(\Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)
b)\(\dfrac{-33}{37}\) và \(\dfrac{-34}{35}\)
Với phân số âm ,phân số nào cũng tử mà khác mẫu ,mẫu nào lớn hơn thì lớn hơn
\(\Rightarrow\dfrac{-33}{37}>\dfrac{-33}{35}\)
c)\(\dfrac{-27}{463}\) và \(\dfrac{-1}{-3}\)
\(\dfrac{-27}{463}< 0;\dfrac{-1}{-3}=\dfrac{1}{3}>0\)
\(\Rightarrow\dfrac{-27}{463}< \dfrac{-1}{-3}\)
21 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
21 tháng 8 2017
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy...
Ta có
2002/2003<1<14/13/
⇒2002/2003<14/13
?????????????????????????