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\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)
\(\Rightarrow x-2009=0\Rightarrow x=2009\)
\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)
\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)
\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2009=0\)
\(\Leftrightarrow x=2009\)
Vậy \(x=2009\)
\(\dfrac{x+4}{2001}+\dfrac{x+3}{2002}=\dfrac{x+2}{2003}+\dfrac{x+1}{2004}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2001}+1\right)+\left(\dfrac{x+3}{2002}+1\right)=\left(\dfrac{x+2}{2003}+1\right)+\left(\dfrac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\dfrac{x+2005}{2001}+\dfrac{x+2005}{2002}-\dfrac{x+2005}{2003}-\dfrac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right)\cdot\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)
Mà \(\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2005=0\Rightarrow x=-2005\)
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
<=>\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
<=>\(\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010
Giải:
Ta có:
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-2=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}-2\)
\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}=\dfrac{x-3-2007}{2007}+\dfrac{x-4-2006}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\Leftrightarrow\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow x=2010\)
Vậy \(x=2010\).
Chúc bạn học tốt!
| \(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\) | ||||
| \(\Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\) | ||||
chuyển vế ta có:
|
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2011}+\dfrac{x-2}{2012}+\dfrac{x-3}{2009}-\dfrac{x-4}{2008}=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\)
\(\Leftrightarrow x-2012=0\Leftrightarrow x=2012\)
Vậy ...
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
=> \(\dfrac{x-1}{2011}-1+\dfrac{x-2}{2010}-1+\dfrac{x-3}{2009}-1=\dfrac{x-4}{2008}-1-2\)
=>\(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}=\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}\)
=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}=0\)=> x - 2012 ( \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\)) = 0
Vì \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\) \(\ge\) 0
=> x - 2012 = 0
=> x = 2012
a)\(A=x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)
Tại \(x=2006\) thì giá trị biểu thức \(A\) là:
\(A=2006^6-2007\cdot2006^5+...-2007\cdot2006+2007\)
\(=2006^6-\left(2006+1\right)\cdot2006^5+...-\left(2006+1\right)\cdot2006+2007\)
\(=2006^6-2006^6+2006^5-...-2006^2-2006+2007\)
\(=-2006+2007=1\)
b)Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Khi đó
\(VT=\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(1\right)\)
\(VP=\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\left(2\right)\)
Từ \((1) và (2)\) ta có điều phải chứng minh
c)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2004\right|+\left|x-1\right|=\left|2004-x\right|+\left|x-1\right|\)
\(\ge\left|2004-x+x-1\right|=2003\)
Đẳng thức xảy ra khi \(1\le x\le2004\)
Vậy với \(1\le x\le2004\) thì \(A_{Min}=2003\)
Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Áp dụng vào bài toán \(\left|x-2004\right|+ \left|x-1\right|\ge\left|x-2004+1-x\right|=2003\)
Dấu "=" xảy ra khi \(\left(x-2004\right)\left(1-x\right)\ge0\)
.....
Giải:
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Leftrightarrow\dfrac{x+4}{2008}+\dfrac{x+3}{2009}+2=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}+2\)
\(\Leftrightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Leftrightarrow\dfrac{x+4+2008}{2008}+\dfrac{x+3+2009}{2009}=\dfrac{x+2+2010}{2010}+\dfrac{x+1+2011}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên \(x+2012=0\)
\(\Leftrightarrow x=0-2012\)
\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\).
Chúc bạn học tốt!
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên:
\(x+2012=0\Rightarrow x=-2012\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15