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\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
Với \(x-2004\ne0\)
\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)
Với \(x-2004=0\)
\(\Rightarrow x=2004\)
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
\(\Leftrightarrow x-2000=0\).Do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Leftrightarrow x=2000\)
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+3=3\)
\(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
Mà: \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Rightarrow\)x-2000=0
Vậy : x=2000
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)
\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)
\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x-2000=0\) ( 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
\(\frac{x}{2000}\)- 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
=> \(x-2000=0\)<=> \(x=2000\)
2.
pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0
<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0
<=> (x-2000).(1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004) = 0
<=> x-2000=0 ( vì 1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004 > 0 )
<=> x=2000
Tk mk nha
1.
a, = (2x-1)^2-2.(2x-1)+1-4
= (2x-1-1)^2-4
= (2x-2)^2-4
= (2x-2-2).(2x-2+2)
= 2x.(2x-4)
b, = [x.(x+3)].[(x+1).(x+2)]
= (x^2+3x).(x^2+3x+1)-8
= (x^2+3x+1)^2-1-8
= (x^2+3x+1)^2-9
= (x^2+3x+1-3).(x^2+3x+1+3)
= (x^2+3x-2).(x^2+3x+4)
= ((x+1).(x+3).(x^2+3x-2)
Tk mk nha
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Rightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Rightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Rightarrow x-2000=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000
Có : \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Leftrightarrow\)\(\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Leftrightarrow\) \(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Leftrightarrow\) \(\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
\(\Leftrightarrow\) \(x-2000=0\)
( vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)\ne0\) )
\(\Leftrightarrow\) \(x=2000\)
Vậy x = 2000 thì phương trình \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
Ta có :
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy ...
Ta có: \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
<=> \(\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)
<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0
<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)
<=> x = 2004
Vậy S = {2004}
đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)
\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)
=> x=1