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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
a, Bạn cộng mỗi tỉ số với 1 rồi chuyển vế phải sang vế trái, ta được:
(x+2016)(1/2011 +1/2012 -1/2013 -1/2014) =0
Ta thấy thừa số thứ hai lớn hơn 0 nên x+2016=0
Vậy x=-2016
b, Bạn chuyển vế phải sang vế trái, ta có:
(5x-1,45)(1/6 +1/7 +1/8 -1/9 +1/10)=0
Thừa số thứ 2 lớn hơn 0 do đó: 5x -1,45 =0
5x =1,45
x =0,29
Vậy x =0,29
Mong bạn hiểu cách giải của mình.
Chúc bạn học tốt.
\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)
\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
Study well
=\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)\)
=\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2106}{2013}-\frac{x-2016}{2012}\)
=\(\left(x-2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\)
Mà: \(\frac{1}{2012}>\frac{1}{2015}\) và \(\frac{1}{2014}< \frac{1}{2013}\)
=>\(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\) khác \(0\)
Nên: \(x-2016=0\)
=>\(x=2016\)
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)
\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
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 \(x-2011=0\)
\(x=2011\)
c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)
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