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1: \(\Leftrightarrow x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+2008\right)=0\)

hay \(x\in\varnothing\)

2: \(x^4+x^2+6x-8=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

hay \(x\in\left\{1;-2\right\}\)

25 tháng 2 2020

\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\\ \Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\\ \Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\\ \Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\\ \Leftrightarrow x+2010=0\\ \Leftrightarrow x=-2010\)

Vậy pt có tập nghiệm \(S=\left\{-2010\right\}\)

31 tháng 7 2018

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)

\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)

\(\Leftrightarrow x+2010=0\)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)

=> x=-2010

vậy.....

8 tháng 5 2020

\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)  ( có lẽ đề như này ) 

\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)

\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)

\(\Leftrightarrow x=2014\)

...

Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)

\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)

\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow3.18=x^2+4x+7x+28\)

\(\Leftrightarrow x^2-2x+13x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)

6 tháng 3 2017

đề ko có vấn đề nhỉ?

7 tháng 3 2017

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

1 tháng 2 2019

a) \(0,25x^3+x^2+x=0\)

\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)

\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)

\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy....

b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)

\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

\(\Rightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy....

30 tháng 1 2019

\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)

\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)

30 tháng 1 2019

\(\text{Giải}\)

\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)

5 tháng 7 2017

b) \(\dfrac{x^2+2\cdot x+2}{x+1}>\dfrac{x^2+4\cdot x+5}{x+2}-1\)

\(\Leftrightarrow\dfrac{x^2+2\cdot x+2}{x+1}-\dfrac{x^2+4\cdot x+5}{x+2}+1>0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x+1\right)\left(x^2+4x+5\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+4x^2+5x+x^2+4x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+5x^2+9x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-x^3-5x^2-9x-5+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{0+0+1}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)>0\)

\(\left\{{}\begin{matrix}x+1>0\\x+2>0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1< 0\\x+2< 0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x>-1\\x>-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x< -1\\x< -2\end{matrix}\right.\)