x-3/2014+x-2/2015=X-2015/2=x-2014/3 giúp mình giải với
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21 tháng 3 2021
Sửa đề: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
Ta có: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
\(\Leftrightarrow\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}+1=\dfrac{x+2001}{15}+1+\dfrac{x+2014}{2}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}=\dfrac{x+2016}{15}+\dfrac{x+2016}{2}\)
\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}-\dfrac{x+2016}{15}-\dfrac{x+2016}{2}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\ne0\)
nên x+2016=0
hay x=-2016
Vậy: S={-2016}
LD
2
24 tháng 3 2021
\(\dfrac{x-2014}{4}+\dfrac{x-2015}{3}=\dfrac{x-13}{2005}+\dfrac{x-14}{2004}\)
<=>\(\left(\dfrac{x-2014}{4}-1\right)+\left(\dfrac{x-2015}{3}-1\right)=\left(\dfrac{x-13}{2005}-1\right)+\left(\dfrac{x-14}{2004}-1\right)\)
<=>\(\dfrac{x-2018}{4}+\dfrac{x-2018}{3}=\dfrac{x-2018}{2005}+\dfrac{x-2018}{2004}\)
<=>\(\left(x-2018\right).\left[\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2005}-\dfrac{1}{2004}\right]=0\)
<=> \(x-2018=0\)
=>x=2018
Vậy S= {2018}
Chúc bạn học tốt!
#Yuii
thank
BT
1
MH
3 tháng 3 2016
x = 2014 => x + 1 = 2015
=> f(2014) = x2014 - (x + 1).x2013 + (x + 1).x2012 - ... - (x + 1).x + x + 1
= x2014 - x2014 - x2013 + x2013 + x2012 - ... - x2 - x + x + 1
= 1
NH
2
7 tháng 5 2017
\(\dfrac{x-5}{2012}+\dfrac{x-4}{2013}=\dfrac{x-3}{2014}+\dfrac{x-2}{2015}\)
\(\Rightarrow\left(\dfrac{x-5}{2012}-1\right)+\left(\dfrac{x-4}{2013}-1\right)=\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}=\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)
Vậy x = 2017
PH
2
7 tháng 2 2020
(x-1)/2015 + x/2014 + 1/503 - (x-3)/2013 - x/2012 - 1/1007 =0
(x-2016)/2015 + (x-2016)/2014 - (x-2016)/2012 - (x-2016)/2013 = 0
(x-2016) ( 1/2015 + 1/2016 - 1/2013 - 1/2012) = 0
Mà 1/2015 + 1/2016 - 1/2013 - 1/2012 khác 0
Suy ra x -2016=0
x=2016
Chỗ nào thắc mắc nhớ hỏi mik nhe!
PH
0
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-2015}{2}-1+\dfrac{x-2014}{3}-1\)
⇔ \(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2}+\dfrac{x-2017}{3}\)
⇔ \(\left(x-2017\right)\)\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
Do : \(\dfrac{1}{2014}< \dfrac{1}{2015}< \dfrac{1}{2}< \dfrac{1}{3}\)
⇒ \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}< 0\)
⇔ x - 2017 = 0
⇔ x = 2017
KL....