Giải phương trình: \(\frac{x}{2017}+\frac{x+1}{2016}=\frac{x+2}{2015}+\frac{x+3}{2014}\)
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9 tháng 3 2019
mk ko chép lại đề nha:
\(\Rightarrow\)\(\frac{x-2}{2017}\)\(-1+\frac{x-3}{2016}\)\(-1=\frac{x-4}{2015}\)\(-1+\frac{x-5}{2014}\)\(-1\)
\(\Rightarrow\)\(\frac{x-2-2017}{2017}\)\(+\frac{x-3-2016}{2016}\)\(=\frac{x-4-2015}{2015}\)\(+\frac{x-5-2014}{2014}\)
\(\Rightarrow\)\(\frac{x-2019}{2017}\)\(+\frac{x-2019}{2016}\)\(-\frac{x-2019}{2015}\)\(-\frac{x-2019}{2014}\)\(=0\)
\(\Rightarrow\)\(\left(x-2019\right)\)\(\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)\)\(=0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-2019=0\\\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}=0\left(voli\right)\end{cases}}\)
\(\Rightarrow\)\(x-2019=0\)
\(\Rightarrow\)\(x=-2019\)
Chỗ mình nghi voli là vô lí nha
chúc bạn học tốt
KN
11 tháng 2 2020
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)
\(\Rightarrow x+10=0\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)
\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0
Vậy x = 2012
11 tháng 2 2020
a, (x+1)/9 +1 + (x+2)/8 = (x+3)/7 + 1 + (x+4)/6 + 1
<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6
<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0
vì 1/9 +1/8 -1/7 - 1/6 khác 0
=> x+10=0
=> x=-10
HX
3
22 tháng 3 2018
Ta có :
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=2^2-4\)
\(\Leftrightarrow\)\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=4-4\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
22 tháng 3 2018
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=0\)
\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=0\)
\(\left(x-2018\right).(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014})=0\)
\(x-2018=0\left(Vì\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\right)\)
\(\Rightarrow x=2018\)
PH
4
21 tháng 6 2017
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
LM
21 tháng 6 2017
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
VT
0
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
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