Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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,\(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
<=> \(\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)
<=>\(\dfrac{2-x}{2001}+1=\left(\dfrac{1-x}{2002}+1\right)+\left(\dfrac{-x}{2003}+1\right)\)
<=>\(\dfrac{2003-x}{2001}=\dfrac{2003-x}{2002}+\dfrac{2003-x}{2003}\)
<=>\(\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
<=> \(\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
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=\(\left\{50\right\}\)
\(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+1-\frac{x}{2003}\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=9\Leftrightarrow2003-x=0\Leftrightarrow x=2003\)
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 \(x>0\)
a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
=>x+36=0
=>x=-36
b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)
=>x-2004=0
=>x=2004
h.
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)
Suy ra: 2004 - x = 0
Vậy x = 2004
\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )
\(\Leftrightarrow x=23\)
Vậy pt có tập nghiệm S = { 23 }
\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { 100 }
\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\)
\(\Leftrightarrow x=300\)
Vậy pt có tập nghiệm S = { 300 }
\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Vậy pt có tập nghiệm S = { 10 }
\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
\(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
Vậy pt có tập nghiệm S = { 2004 }
\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { -100 }
<