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G
11 tháng 5 2018
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 ...
JD
11 tháng 5 2018
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>3\)
<=>\(\left(\dfrac{x+1}{2953}+1\right)+\left(\dfrac{x+953}{2001}+1\right)+\left(\dfrac{x+2950}{4}+1\right)>0\)
<=>\(\dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\)
<=>\(\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\)
Vì \(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}>0\) nên \(x+2954>0\) <=> \(x>-2954\)
KL: ...
7 tháng 8 2017
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\\ \dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>-3+3\\ \dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\\ \left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\\ x+2954>0\\ x>-2954\)
7 tháng 8 2017
a.2mx=0 <=> mx=0
•nếu m=0 thì nghiệm đúng với mọi x
•nếu \(m\ne0\) thì nghiệm đúng với x=0
MV
12 tháng 8 2017
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 \(x>0\)
1 tháng 4 2017
(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
22 tháng 4 2021
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: \(S=\varnothing\)
22 tháng 4 2021
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}
30 tháng 3 2017
\(\dfrac{x-4}{2001}\)- 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
30 tháng 3 2017
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
LT
2
7 tháng 2 2021
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 ...
7 tháng 2 2021
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}
LT
1
7 tháng 2 2021
ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x+1}{X-1}-\dfrac{x-1}{X+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)(loại)
Vậy phương trình vô nghiệm
LT
2
7 tháng 2 2021
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}
TG
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}>-2\)
\(\Leftrightarrow\dfrac{x+1}{2953}+1+\dfrac{x+953}{2001}+1>-2+1+1\)
\(\Leftrightarrow\dfrac{x+2954}{2953}+\dfrac{x+2955}{2001}>0\)
\(\Leftrightarrow\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}\right)>0\)
\(\Leftrightarrow x+2954>0\\ \Leftrightarrow x>-2954\)
Vậy.......