giải bất phương trình sau:
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\)
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1
5 tháng 4 2020
\(\frac{x+1}{2953}+\frac{x+953}{2001}+\frac{x+2950}{4}>-3\)
\(\Leftrightarrow\frac{x+1}{2953}+\frac{x+953}{2001}+\frac{x+2950}{4}+3>0\)
\(\Leftrightarrow\frac{x+1}{2953}+1+\frac{x+953}{2001}+1+\frac{x+2950}{4}+1>0\)
\(\Leftrightarrow\frac{x+1+2953}{2953}+\frac{x+953+2001}{2001}+\frac{x+2950+4}{4}>0\)
\(\Leftrightarrow\frac{x+2954}{2953}+\frac{x+2954}{2001}+\frac{x+2954}{4}>0\)
\(\Leftrightarrow\left(x+2954\right)\left(\frac{1}{2953}+\frac{1}{2001}+\frac{1}{4}\right)>0\)
Vì \(\frac{1}{2953}+\frac{1}{2001}+\frac{1}{4}>0\)
Nên \(x+2954>0\)
\(\Leftrightarrow x>-2954\)
Vậy .........
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
HS
8 tháng 8 2018
\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}>-2\)
\(\Leftrightarrow\dfrac{x+1}{2953}+1+\dfrac{x+953}{2001}+1>-2+1+1\)
\(\Leftrightarrow\dfrac{x+2954}{2953}+\dfrac{x+2955}{2001}>0\)
\(\Leftrightarrow\left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}\right)>0\)
\(\Leftrightarrow x+2954>0\\ \Leftrightarrow x>-2954\)
Vậy.......
HN
5 tháng 4 2022
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
KV
5 tháng 4 2022
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)
16 tháng 4 2023
a: =>x^2-8x+16<x^2-8x
=>16<0(loại)
b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)
=>x+1/2>=5/3x-1
=>-2/3x>=-3/2
=>x<=3/2:2/3=9/4
c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)
=>21-3x<=8x-16
=>-11x<=-37
=>x>=37/11
10 tháng 5 2023
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
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
22 tháng 5 2017
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x+\left(2x-1\right)}{6}=\dfrac{24-2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x+2x=24+1\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{25}{8}\)
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)+3\left(x-1\right)}{12}=\dfrac{12-8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow9\left(x-1\right)+8\left(x-1\right)=12\)
\(\Leftrightarrow17\left(x-1\right)=12\)
\(\Leftrightarrow17x-17=12\)
\(17x=12+17\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy phương trình có một nghiệm là x = \(\dfrac{29}{17}\)
c) \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-\dfrac{1-x}{2002}-\dfrac{\left(-x\right)}{2003}=1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1-\dfrac{1-x}{2002}-1-\dfrac{\left(-x\right)}{2003}-1=1+1-1-1\)
\(\Leftrightarrow\dfrac{2-x}{2001}+\dfrac{2001}{2001}-\dfrac{1-x}{2002}-\dfrac{2002}{2002}-\dfrac{\left(-x\right)}{2003}-\dfrac{2003}{2003}=0\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\)
\(\Leftrightarrow-x=-2003\)
\(\Leftrightarrow x=2003\)
Vậy phương trình có một nghiệm là x = 2003
29 tháng 5 2017
a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}=\dfrac{24}{6}-\dfrac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow4x+2x+2x=1+24\)
\(\Leftrightarrow8x=25\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
Vậy S={\(\dfrac{25}{8}\)}
b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}+\dfrac{3\left(x-1\right)}{12}=\dfrac{12}{12}-\dfrac{8\left(x-1\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)
\(\Leftrightarrow6x-6+3x-3=12-8x+8\)
\(\Leftrightarrow6x+3x+8x=6+3+12+8\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy S={\(\dfrac{29}{17}\)}
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: ...