Giải phương trình :
\(\frac{2015}{2016}\).x + \(\frac{2016}{2017}\).x + \(\frac{2017}{2018}\).x = \(\frac{2018}{2019}\).x
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L
1
24 tháng 8 2019
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
DX
0
OM
5 tháng 4 2020
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
HT
0
JK
23 tháng 6 2020
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(\Leftrightarrow\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-3}{2018}-1\)
\(\Leftrightarrow\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=2020\)
PN
23 tháng 6 2020
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(< =>\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-2}{2018}-1\)
\(< =>\frac{x-5-2015}{2015}+\frac{x-4-2016}{2016}=\frac{x-3-2017}{2017}+\frac{x-2-2018}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)
\(< =>\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
\(< =>x-2020=0< =>x=2020\)
NL
2
29 tháng 9 2016
Mình chắc 50% thôi
Nếu x = 0 thì phù hợp đó
vì 0/2015=0/2016=0/2017=0
mà 0+0+0=0/2018=0
29 tháng 9 2016
bn cứ chắc 100% đi, là bài violympic đó,tui giúp bn chắc 100%
x( 1/2015 + 1/2016 + 1/2017 - 1/ 2018) = 0
=> x = 0
Ta có \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)
<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)
<=>x\(\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Vì \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\) không thể bằng 0
Vậy x=0
Ta có 1 nghiệm thỏa mãn S=\(\left\{0\right\}\)