x-5/1007+x-3/504=x-1/2018+x-4/403
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NT
1
26 tháng 4 2020
\(\frac{x-3}{504}+\frac{x-5}{1007}=\frac{x-1}{2018}+\frac{x-4}{403}\)
<=> \(\frac{x-3}{504}-4+\frac{x-5}{1007}-2=\frac{x-1}{2018}-1+\frac{x-4}{403}-5\)
<=> \(\frac{x-2019}{504}+\frac{x-2019}{1007}=\frac{x-2019}{2018}+\frac{x-2019}{403}\)
<=> \(\left(x-2019\right)\left(\frac{1}{504}+\frac{1}{1007}-\frac{1}{2018}-\frac{1}{403}\right)=0\)
<=> x - 2019 = 0
( vì \(\frac{1}{504}+\frac{1}{1007}-\frac{1}{2018}-\frac{1}{403}\ne0\))
<=> x = 2019
vậy x = 2019.
16 tháng 3 2018
\(5-\frac{x}{2010}+4-\frac{x}{2011}+3-\frac{x}{2012}=6-\frac{x}{2009}+1-\frac{x}{1007}.\)
\(\left(5+4+3\right)-x.\frac{1}{2010}-x.\frac{1}{2011}-x\frac{1}{2012}=\left(6+1\right)-x.\frac{1}{2009}-x\frac{1}{1007}\)
\(12-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=7-x.\left(\frac{1}{2009}+\frac{1}{1007}\right)\)
\(-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)+x.\left(\frac{1}{2009}+\frac{1}{1007}\right)=7-12\)
\(x.\left(\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}\right)=-5\)
\(x=\frac{-5}{\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}}\)
SR
3
ML
20 tháng 7 2017
Sửa đề :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2018-1\)
\(\Leftrightarrow x=2017\)
Vậy ...
DP
20 tháng 7 2017
Sửa đề \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2017\)
24 tháng 6 2018
\(\frac{x-18}{2018}+\frac{x-14}{1007}+\frac{x-13}{671}=-6\)
\(\Rightarrow\frac{x-18}{2018}+1+\frac{x-14}{1007}+2+\frac{x-13}{671}+3=-6+6\)
\(\Rightarrow\frac{x-2000}{2028}+\frac{x-2000}{1007}+\frac{x-2000}{671}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{1007}+\frac{1}{671}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{1007}+\frac{1}{671}\ne0\)
=> x - 2000 = 0
=> x = 2000
6 tháng 6 2018
Lập bảng xét dấu nhé :
| x \(\frac{1}{3}\) 2015 |
| x - 2015 - - 0 + |
| 3x - 1 - 0 + + |
Th 1 : \(x< \frac{1}{3}\) pt trở thành : \(2015-x+1-3x=0\)
\(\Leftrightarrow2016-4x=0\)
\(\Leftrightarrow4x=2016\)
\(\Leftrightarrow x=504\) (loại)
Th2 : \(\frac{1}{3}\le x< 2015\) pt trở thành : \(2015-x+3x-1=0\)
<=> 2014 - 2x = 0
<=> 2x = 2014
<=> x = 1007 (t/m)
Th3 : \(x\ge2015\) thì pt trở thành : \(x-2015+3x-1=0\)
<=> 4x - 2016 = 0
<=> 4x = 2016
<=> x = 504
Vậy ...................................
Đáp án C nhé !
4 tháng 8 2017
\(A=\left|2x-\dfrac{1}{3}\right|+1007\)
\(\left|2x-\dfrac{1}{3}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{3}\right|+1007\ge1007\)
Dấu "=" xảy ra khi:
\(\left|2x-\dfrac{1}{3}\right|=0\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\Rightarrow MIN_A=1007\) khi \(x=\dfrac{1}{6}\)
B tương tự
\(C=\left|2018-x\right|+\left|2017-x\right|\)
\(C=\left|2018-x\right|+\left|x-2017\right|\)
Áp dụng BĐT:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow C\ge\left|2018-x+x-2017\right|\)
\(C\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2018-x\ge0\Rightarrow x\le2018\\x-2017\ge0\Rightarrow x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}2018-x< 0\Rightarrow x>2018\\x-2017< 0\Rightarrow x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2017\le x\le2018\)
D tương tự