Giải:

\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\)

\(\Leftrightarrow\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>0\)

\(\Leftrightarrow\dfrac{x+1}{2953}+1+\dfrac{x+953}{2001}+1+\dfrac{x+2950}{4}+1>0\)

\(\Leftrightarrow\dfrac{x+1+2953}{2953}+\dfrac{x+953+2001}{2001}+\dfrac{x+2950+4}{4}>0\)

\(\Leftrightarrow\dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\)

\(\Leftrightarrow\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\)

Vì \(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}>0\)

Nên \(x+2954>0\)

\(\Leftrightarrow x>-2954\)

Vậy ...

\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>3\)

<=>\(\left(\dfrac{x+1}{2953}+1\right)+\left(\dfrac{x+953}{2001}+1\right)+\left(\dfrac{x+2950}{4}+1\right)>0\)

<=>\(\dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\)

<=>\(\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\)

Vì \(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}>0\) nên \(x+2954>0\) <=> \(x>-2954\)

KL: ...

\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\\ \dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>-3+3\\ \dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\\ \left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\\ x+2954>0\\ x>-2954\)

a.2mx=0 <=> mx=0

•nếu m=0 thì nghiệm đúng với mọi x

•nếu \(m\ne0\) thì nghiệm đúng với x=0

\(\frac{x+1}{2953}+\frac{x+953}{2001}+\frac{x+2950}{4}>-3\)

\(\Leftrightarrow\frac{x+1}{2953}+\frac{x+953}{2001}+\frac{x+2950}{4}+3>0\)

\(\Leftrightarrow\frac{x+1}{2953}+1+\frac{x+953}{2001}+1+\frac{x+2950}{4}+1>0\)

\(\Leftrightarrow\frac{x+1+2953}{2953}+\frac{x+953+2001}{2001}+\frac{x+2950+4}{4}>0\)

\(\Leftrightarrow\frac{x+2954}{2953}+\frac{x+2954}{2001}+\frac{x+2954}{4}>0\)

\(\Leftrightarrow\left(x+2954\right)\left(\frac{1}{2953}+\frac{1}{2001}+\frac{1}{4}\right)>0\)

Vì \(\frac{1}{2953}+\frac{1}{2001}+\frac{1}{4}>0\)

Nên \(x+2954>0\)

\(\Leftrightarrow x>-2954\)

Vậy .........

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)

(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)

Vậy x>9 thì (x-5)(x-9)>0

có

\(\dfrac{x-5}{x-8}>2\\ < =>x-5>2\left(x-8\right)\\ < =>x-5>2x-16\\ < =>-x>-11\\ < =>x< 11\)

vậy nghiệm của bpt là x<11

a/

\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)

\(\Leftrightarrow\dfrac{x+2014-2011}{2011}+\dfrac{x+2014-2013}{2013}\ge\dfrac{x+2014-2004}{2004}+\dfrac{x+2014-2001}{2001}\)

\(\Leftrightarrow-1+\dfrac{x+2014}{2011}-1+\dfrac{x+2014}{2013}\ge-1+\dfrac{x+2014}{2004}-1+\dfrac{x+2014}{2001}\)

\(\Leftrightarrow\dfrac{x+2014}{2011}+\dfrac{x+2014}{2013}-2\ge\dfrac{x+2014}{2004}+\dfrac{x+2014}{2001}-2\)

\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)

\(\Leftrightarrow\dfrac{1}{2011}+\dfrac{1}{2013}>\dfrac{1}{2004}+\dfrac{1}{2001}\) hoặc \(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)

(với mọi x>0) \(\Leftrightarrow x=2014\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005

x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4

<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+

(x-2002/3 -1)+(x-2001/4 -1) =0

<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-

x-2005/3- x-2005/4 =0

<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0

<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)

<=>x=2005

Vậy pt có nghiệm là x=2005

\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)

\(\Leftrightarrow\dfrac{x-4}{2001}-1+\dfrac{x-3}{2002}-1+\dfrac{x-2}{2003}-1=\dfrac{x-2003}{2}-1+\dfrac{x-2002}{3}-1+\dfrac{x-2001}{4}-1\)

\(\Leftrightarrow\dfrac{x-2005}{2001}+\dfrac{x-2005}{2002}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2}-\dfrac{x-2005}{3}-\dfrac{x-2005}{4}=0\)

\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\right)=0\)

\(\Leftrightarrow x-2005=0\)

\(\Leftrightarrow x=2005\)

Vậy nghiệm của PT là \(x=2005\)

ĐKXĐ : \(x\ne\pm1\)

PT : \(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}=\dfrac{x+1-\left(x+2\right)\left(x-1\right)}{x-1}\)

\(\Leftrightarrow\dfrac{1-x^2}{x+1}=1-x=\dfrac{3-x^2}{x-1}\)

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

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy ...

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x-1}{x+1}-\dfrac{x^2+x-2}{x+1}=\dfrac{x+1}{x-1}-x-2\)

\(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)

\(\Leftrightarrow\dfrac{-x^2+1}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)

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

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

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

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

Suy ra: \(-\left(x^2-2x+1\right)-x-1+x^2-x+2x-2=0\)

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

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(nhận)

Vậy: S={2}

\(DKXD:x\ne\pm1\\ pt\Rightarrow2x^2+x+1+2x-2=x^2-1\\ \Leftrightarrow x^2+3x=0\\ \Leftrightarrow x\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\left(N\right)\\ \Rightarrow S=\left\{0;-3\right\}\)