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\(\dfrac{x}{2000}+\dfrac{x+1}{2001}+\dfrac{x+2}{2002}+\dfrac{x+3}{2003}+\dfrac{x+4}{2004}=5\)
\(\Leftrightarrow\dfrac{x}{2000}-1+\dfrac{x+1}{2001}-1+\dfrac{x+2}{2002}-1+\dfrac{x+3}{2003}-1+\dfrac{x+4}{2004}-1=0\)
\(\Leftrightarrow\dfrac{x-2000}{2000}+\dfrac{x-2000}{2001}+\dfrac{x-2000}{2002}+\dfrac{x-2000}{2003}+\dfrac{x-2000}{2004}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)
Mà \(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}>0\)
\(\Leftrightarrow x-2000=0\Leftrightarrow x=2000\)
Vậy x = 2000
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
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={\(\dfrac{29}{17}\)}
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
a,\(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
<=> \(\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)
<=>\(\dfrac{2-x}{2001}+1=\left(\dfrac{1-x}{2002}+1\right)+\left(\dfrac{-x}{2003}+1\right)\)
<=>\(\dfrac{2003-x}{2001}=\dfrac{2003-x}{2002}+\dfrac{2003-x}{2003}\)
<=>\(\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
<=> \(\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
=> \(2003-x=0\)
=> \(x=2003\)
Vậy : S = \(\left\{2003\right\}\)
b, \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}+\dfrac{2x-5}{95}=\dfrac{2x-6}{94}\)
<=> \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}=\dfrac{2x-6}{94}-\dfrac{2x-5}{95}\)
<=> \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}-2=\dfrac{2x-6}{94}-\dfrac{2x-5}{95}-2\)
<=> \(\left(\dfrac{2x-3}{97}-1\right)-\left(\dfrac{2x-4}{96}-1\right)=\left(\dfrac{2x-6}{94}-1\right)-\left(\dfrac{2x-5}{95}-1\right)\)
<=>\(\dfrac{2x-100}{97}-\dfrac{2x-100}{96}=\dfrac{2x-100}{94}-\dfrac{2x-100}{95}\)
<=> \(\dfrac{2x-100}{97}-\dfrac{2x-100}{96}-\dfrac{2x-100}{94}+\dfrac{2x-100}{95}=0\)
<=> \(\left(2x-100\right)\left(\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{94}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{94}+\dfrac{1}{95}\ne0\)
=>\(2x-100=0\)
=> \(2x=100\)
=>\(x=50\)
Vậy: S=\(\left\{50\right\}\)
\(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+1-\frac{x}{2003}\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=9\Leftrightarrow2003-x=0\Leftrightarrow x=2003\)
\(\dfrac{x+101}{2001}+\dfrac{x+99}{2003}=\dfrac{x+100}{2002}+\dfrac{x+98}{2004}\)
\(\Leftrightarrow\left(\dfrac{x+101}{2001}+1\right)+\left(\dfrac{x+99}{2003}+1\right)=\left(\dfrac{x+100}{2002}+1\right)+\left(\dfrac{x+98}{2004}+1\right)\)
\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}=\dfrac{x+2102}{2002}+\dfrac{x+2102}{2004}\)
\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}-\dfrac{x+2102}{2002}-\dfrac{x+2102}{2004}=0\)
\(\Leftrightarrow\left(x+2102\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\right)=0\)
Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\ne0\)
\(\Rightarrow x+2102=0\)
\(\Rightarrow x=-2102\)
\(\Rightarrow S=\left\{-2102\right\}\)
a, \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1=\left(\dfrac{1-x}{2002}+1\right)+\left(\dfrac{-x}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{2003-x}{2001}=\dfrac{2003-x}{2002}+\dfrac{2003-x}{2003}\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow2003-x=0\)
\(\Rightarrow x=2003\)
Vậy : \(s=\left\{2003\right\}\)
b, \(\dfrac{x-5}{100}+\dfrac{x-4}{101}=\dfrac{x-100}{5}+\dfrac{x-101}{4}\)
\(\Leftrightarrow\dfrac{x-5}{100}+\dfrac{x-4}{101}-2=\dfrac{x-100}{5}+\dfrac{x-101}{4}-2\)
\(\Leftrightarrow\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}=\dfrac{x-105}{5}+\dfrac{x-105}{4}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}-\dfrac{x-105}{5}-\dfrac{x-105}{4}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
Vì \(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)
\(\Rightarrow x-105=0\)
\(\Rightarrow x=105\)
Vậy : \(s=\left\{105\right\}\)
\(a,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\)haizzz bạn cộng mỗi hạng tử ở mỗi vế cho một. Chuyển vế và giải ra x=2003
b, Tương tự bạn -1 cho mỗi vế. GIải phương trình đc x=105
\(\dfrac{x-4}{2001}\)- 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