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\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
Lâp bảng xét dấu
2016 2017
x-2016 _ 0 + +
x-2017 _ _ 0 +
Nếu x<2016 thì |x-2016|=2016-x,|x-2017|=2017-x
Ta có 2016-x+2017-x=2018
4033-2x=2018
2x=2015
x=1007,5
Nếu 2016<=x<=2017thif |x-2016|=x-2016;|x-2017|=2017-x
Ta có x-2016+2017-x=2018
ox+1=2018
0x=2017 (vô lí)
Nếu x>=2017 thi |x-2016|=x-2016;|x-2017|=x-2017
Ta có x-2016+x-2017=2018
2x-4033=2018
2x=6051
x=3025,5
Vậy x=1007,5 hoăc x=3025,5
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
Tìm x biết:
\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)
Giải:Ta có:\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)
\(\Rightarrow\frac{x}{2018}+1+\frac{x+1}{2017}+1+\frac{x+2}{2016}+1+\frac{x+3}{2015}+1=0\)
\(\Rightarrow\frac{x+2018}{2018}+\frac{x+2018}{2017}+\frac{x+2018}{2016}+\frac{x+2018}{2015}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\) vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}>0\)
\(\Rightarrow x=-2018\)
Vậy x=-2018 thỏa mãn
x2018 +x+12017 +x+22016 +x+32015 =−4
⇒x2018 +1+x+12017 +1+x+22016 +1+x+32015 +1=0
⇒x+20182018 +x+20182017 +x+20182016 +x+20182015 =0
⇒(x+2018)(12018 +12017 +12016 +12015 )=0
⇒x+2018=0 vì 12018 +12017 +12016 +12015 >0
⇒x=−2018
Vậy x=-2018 thỏa mãn
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)
\(x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)=0\)
Mà \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)\ne0\)
=> x = 0
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)
\(\Leftrightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1=\frac{x+2}{2016}+1+\frac{x+1}{2017}+1\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
Vì: \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)