\(\frac{3x+1}{2002}+1=\frac{2-3x}{2003}+\frac{4x+2}{2001}\)
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22 tháng 1 2016
\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2001}{2002}.\frac{2002}{2003}=\frac{1.2.3.....2001.2002}{2.3.4.....2002.2003}=\frac{1}{2003}\)
QA
2
10 tháng 8 2016
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2002}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\) \(x=2003\)
10 tháng 8 2016
↔ \(\frac{2-x}{2001}+1\)\(=\left(\frac{1-x}{2002}+1\right)+\left(\frac{x}{2003}+1\right)\)
↔ \(\frac{2003-x}{2001}\) \(=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
↔ \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
↔ x = 2003
27 tháng 2 2019
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
YN
17 tháng 2 2020
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)
\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)
\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x-2000=0\) ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )
\(\Leftrightarrow x=2000\)
Vậy x = 2000
Đây là cách của lớp 7 nha
@@ Học tốt
17 tháng 2 2020
\(\frac{x}{2000}\)- 1+\(\frac{x+1}{2001}\)-1+\(\frac{x+2}{2002}\)-1+\(\frac{x+3}{2003}\)-1=0
<=>\(\frac{x-2000}{2000}\)+ \(\frac{x-2000}{2001}\)+ \(\frac{x-2000}{2002}\)+ \(\frac{x-2000}{2003}\)=0
<=>\(\left(x-2000\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)=0
Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0
=> \(x-2000=0\)<=> \(x=2000\)
10 tháng 11 2016
\(\frac{y-1}{2004}+\frac{y-2}{2003}-\frac{y-3}{2002}=\frac{y-4}{2001}\)
\(\frac{y-1}{2004}-1+\frac{y-2}{2003}-1-\frac{y+3}{2002}+1=\frac{y-4}{2001}-1\)
\(\frac{y-2005}{2004}+\frac{y-2005}{2003}-\frac{y-2005}{2002}=\frac{y-2005}{2001}\)
\(\frac{y-2005}{2001}+\frac{y-2005}{2002}-\frac{y-2005}{2003}-\frac{y-2005}{2004}=0\)
\(\left(y-2005\right).\left(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Dễ thấy: \(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}>0\)
=> y - 2005 = 0
=> y = 2005