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\(\dfrac{2-x}{2007}\) - 1 = \(\dfrac{1-x}{2008}\) - \(\dfrac{x}{2009}\)
<=> \(\dfrac{2-x}{2009}\) +1 -1 +1 = \(\dfrac{1-x}{2008}\) +1 - \(\dfrac{x}{2009}\) +1
<=> \(\dfrac{2-x+2007}{2007}\) = \(\dfrac{1-x+2008}{2008}\) + \(\dfrac{-x+2009}{2009}\)
<=> \(\dfrac{2009-x}{2007}\) = \(\dfrac{2009-x}{2008}\) + \(\dfrac{2009-x}{2009}\)
<=> (2009-x)(\(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) - \(\dfrac{1}{2009}\) ) = 0
<=> 2009 -x = 0
hoặc: \(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) -\(\dfrac{1}{2009}\) = 0
Vì \(\dfrac{1}{2007}\) \(\ne\) \(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\)
=> \(\dfrac{1}{2007}\) - (\(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\) ) \(\ne\) 0
=> 2009 -x =0
<=> x =2009
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2009-x}{2007}-2=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}-2\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
\(\Rightarrow2009-x=0\Leftrightarrow x=2009\)
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
\(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\left(l\right)\\x=1\\x=-1\end{matrix}\right.\)
\(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}\left(x+5\right)\\2x-1=-\dfrac{4}{3}\left(x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}x+\dfrac{20}{3}\\2x-1=-\dfrac{4}{3}x-\dfrac{20}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{23}{3}\\\dfrac{2}{3}x=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(l\right)\\x=-\dfrac{17}{10}\left(n\right)\end{matrix}\right.\)
Vậy: \(x=-\dfrac{17}{10}\)
\(C=\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}+...+\dfrac{\sqrt{x+2007}-\sqrt{x+2008}}{-1}\)
\(=-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-...-\sqrt{x+2007}+\sqrt{x+2008}\)\(=-\sqrt{x}+\sqrt{x+2008}\)
\(C=-\sqrt{\sqrt[2007]{2008}}+\sqrt{\sqrt[2007]{2008}+2008}\)
Không chẳng có vấn đề gì cả. có thể sai so với cái đề nào đó "nội hàm nó đúng"
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{-x+4}{2006}+\dfrac{-x-2008}{6}\)
\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).x=\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)\)\(x=\dfrac{\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)}{\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).}\)
Thích thì rút gọn chẳng thích thì kệ nó
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ x+2009=0
⇔ x=-2009
vậy x=-2009 là nghiệm của pt
a) ( x2 + x )2 + 4( x2 + x ) = 12
<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0
<=> ( x2 + x + 2)2 - 16 = 0
<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0
<=> ( x2 + x + 6 )( x2 + x - 2) = 0
Do : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x
=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1) + 2( x - 1) = 0
<=> ( x - 1)( x + 2 ) = 0
<=> x = 1 hoặc : x = - 2
KL.....
b) Kuroba kaito làm rùi nhé 
a)\(\dfrac{2x+1}{x-3}-\dfrac{x}{x+3}=0\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(x+3\right)-\left(x-3\right)x=0\)
\(\Leftrightarrow2x^2+7x+3-x^2+3x=0\)
\(\Leftrightarrow x^2+10x+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5+\sqrt{22}\\x=-5-\sqrt{22}\end{matrix}\right.\)(tm)
b.
\(\left(x-4\right)\left(x-5\right)=12\)
\(\Leftrightarrow x^2-9x+20-12-0\)
\(\Leftrightarrow x^2-9x+8=0\)
\(\Leftrightarrow x^2-8x-x+8=0\)
\(\Leftrightarrow x\left(x-8\right)-\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
Vậy........
Giải:
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\dfrac{2-x}{2007}-1+2=\dfrac{1-x}{2008}-\dfrac{x}{2009}+2\)
\(\Leftrightarrow\dfrac{2-x}{2007}+1=\dfrac{1-x}{2008}+1-\dfrac{x}{2009}+1\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}-\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
Vì \(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\ne0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)
Vậy ...
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\left(\dfrac{2-x}{2007}+1\right)-\left(1+1\right)=\left(\dfrac{1-x}{2008}+1\right)-\left(\dfrac{x}{2009}+1\right)\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}+\dfrac{-x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)