\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
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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
FF
2
22 tháng 4 2020
Bài 1 :
Ta có :
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)
\(+\left(\frac{x+2013}{2011}+1\right)\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+4024=0\)
\(\Rightarrow x=-4024\)
22 tháng 4 2020
Bài 2 :
Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)
=> Phương trình trở thành
\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)
\(\Rightarrow5a^2+3a-8=0\)
\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)
Vì \(a\ge0\Rightarrow a=1\)
\(\Rightarrow x^2+2x+1=1\)
\(x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2,0\right\}\)
16 tháng 2 2020
a, \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
= \(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-1}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
( x + 5)(x - 3) = \(x^2-1\) - 8
x\(^2\) -3x + 5x -15 = \(x^2-9\)
= > \(x^2-x^2\) +2x = 15 - 9
=> 2x = 6
=> x = 3
HP
13 tháng 12 2015
cộng 1 vào mỗi tỉ số ta được:
\(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1+\frac{x+3}{2014}+1=\frac{x+4}{2013}+1+\frac{x+5}{2012}+\frac{x+6}{2011}\)
=>\(\frac{x+1}{2016}+\frac{2016}{2016}+\frac{x+2}{2015}+\frac{2015}{2015}+\frac{x+3}{2014}+\frac{2014}{2014}=\frac{x+4}{2013}+\frac{2013}{2013}+\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+6}{2011}+\frac{2011}{2011}\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}=\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\left(\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\right)=0\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
=>(x+2017).(1/1016+1/2015+1/2014-1/2013-1/2012-1/2011)=0
dễ thấy 1/2016<1/2015<1/2014<1/2013<1/2012<1/2011
=>1/2016+...-1/2011 khác 0
=>x+2017=0
=>x=-2017
nhớ tick
T
17 tháng 9 2018
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
BN
16 tháng 7 2017
\(\Leftrightarrow\frac{x+1}{2009}+\frac{x+1}{2010}+\frac{x+1}{2011}-\frac{x+1}{2012}-\frac{x+1}{2013}-\frac{x+1}{2014}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}=0\end{cases}}\)
mà \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
nên \(x+1=0\)
\(\Leftrightarrow x=-1\)
. Ta có: \(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\) \(\Leftrightarrow\frac{x+1}{2016}+1+\frac{x+3}{2014}+1=\frac{x+5}{2012}+1\frac{x+7}{2010}+1\)
. \(\Leftrightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\) \(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)\)
\(\Leftrightarrow x+2017=0\) \(\Leftrightarrow x=-2017\)
\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2016}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+7}{2010}+1\right)\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}=\frac{x+2017}{2012}+\frac{x+2017}{2010}\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2017=0\)\(\left(Vì\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\ne0\right)\)
\(\Rightarrow x=0-2017\)
\(\Rightarrow x=-2017\)
Vậy x=-2017