Tìm x biết
\(\frac{x+5}{2010}+\frac{x+4}{2011}=\frac{x+2010}{5}+\frac{x+2012}{3}\)
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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 \(x-2011=0\)
\(x=2011\)
NA
6
NA
1
MT
11 tháng 7 2015
\(\frac{x+2}{2012}+\frac{x+3}{2011}=\frac{x+4}{2010}+\frac{x+5}{2009}\)
\(\Rightarrow\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1\)
\(\frac{x+2}{2012}+\frac{2012}{2012}+\frac{x+3}{2011}+\frac{2011}{2011}=\frac{x+4}{2010}+\frac{2010}{2010}+\frac{x+5}{2009}+\frac{2009}{2009}\)
\(\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}\)
\(\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}=0\)
\(\left(x+2014\right)\left(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\ne0\)
nên \(x+2014=0\)
\(x=-2014\)
26 tháng 11 2019
Áp dụng BĐT Cô - si ngược dấu :
\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4\left(x-2010\right)}\le\frac{4+\left(x-2010\right)}{4}\)
\(\Rightarrow\sqrt{x-2010}-1\le\frac{4+\left(x-2010\right)}{4}-1=\frac{x-2010}{4}\)
\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}\le\frac{1}{4}\)
Hoàn toàn tương tự với những phân thức còn lại
\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)
N
2
T
2 tháng 1 2017
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2013}+1\right)+\left(\frac{x+2}{2012}+1\right)=\left(\frac{x+3}{2011}+1\right)+\left(\frac{x+4}{2010}+1\right)\)
\(\Rightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}-\frac{x+2014}{2011}-\frac{x+2014}{2010}=0\)
\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)nên để \(\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Thì x+2014=0
=>x=-2014
2 tháng 1 2017
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
=> \(\frac{x+1+2013}{2013}+\frac{x+2+2012}{2012}=\frac{x+3+2011}{2011}+\frac{x+4+2010}{2010}\)
=> \(\frac{x+2014}{2013}+\frac{x+2014}{2012}=\frac{x+2014}{2011}+\frac{x+2014}{2010}\)
=> \(\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
=> \(x+2014=0\)(do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\))
=> \(x=-2014\)
NT
2
CW
1 tháng 1 2017
\(\frac{x-3}{2013}+\frac{x-4}{2012}=\frac{x-5}{2011}+\frac{x-6}{2010}\)
\(\Leftrightarrow\frac{x-3-2013}{2013}+\frac{x-2-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)(mỗi vế trừ đi 2)
\(\Leftrightarrow\frac{x-2016}{2013}+\frac{x-2016}{2012}-\frac{x-2016}{2011}-\frac{x-2016}{2010}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Mà \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)
\(\Rightarrow x-2016=0\Leftrightarrow x=2016\)
TC
5 tháng 10 2019
Cộng mỗi vế cho 1
Ta có: \(\frac{x-3-2013}{2013}+\frac{x-4-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)
\(=>\left(\frac{x-2016}{2013}+\frac{x-2016}{2012}\right)-\left(\frac{x-2016}{2011}+\frac{x-2016}{2010}\right)=0\)
\(=>\left(x-2016\right).\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)\)
Mà \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\ne0\)
\(=>x-2016=0\\ =>x=2016\)
2 tháng 4 2018
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+\frac{x-4}{2009}+\frac{x-2021}{2}=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+\left(\frac{x-4}{2009}-1\right)+\left(\frac{x-2021}{2}+4\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+\frac{x-2013}{2009}+\frac{x-2013}{2}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\ne0\)
Nên \(x-2013=0\)
\(\Rightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~
TN
7 tháng 3 2017
\(\frac{x-2}{2012}+\frac{x-3}{2011}+\frac{x-4}{2010}+\frac{x-2029}{5}=0\)
\(\Leftrightarrow\frac{x-2}{2012}-1+\frac{x-3}{2011}-1+\frac{x-4}{2010}-1+\frac{x-2029}{5}+3=0\)
\(\Leftrightarrow\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}+\frac{x-2014}{5}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow x-2014=0\).Do \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\ne0\)
\(\Leftrightarrow x=2014\)
Ta có: \(\frac{x+5}{2010}+\frac{x+4}{2011}=\frac{x+2010}{5}+\frac{x+2012}{3}\)
\(\Leftrightarrow\frac{x+5}{2010}+1+\frac{x+4}{2011}+1=\frac{x+2010}{5}+1+\frac{x+2012}{3}+1\)
\(\Leftrightarrow\frac{x+5+2010}{2010}+\frac{x+4+2011}{2011}=\frac{x+2010+5}{5}+\frac{x+2012+3}{3}\)
\(\Leftrightarrow\frac{x+2015}{2010}+\frac{x+2015}{2011}-\frac{x+2015}{5}-\frac{x+2015}{3}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{5}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{5}-\frac{1}{3}\ne0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=0-2015=-2015\)
Vậy \(x=-2015\)