tim x
a. x-10/2010 + x-3/2003 + x-2/2002 =-3
b. x-1009/1001 + x-4/1003 + x+10/1005 =7
K
Khách
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SO
23 tháng 3 2020
\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)=7
⇔\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)-7=0
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔(x-2010)\(\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)\)=0
⇔x-2010=0
⇔x=2010
Vậy x=2010
DT
23 tháng 3 2020
\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
⇔ \(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}-7=0\)
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)\)\(+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\)\(\frac{x+2010-4020}{1005}=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔\(\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
⇔ \(x-2010=0\left(do\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}>0\right)\)
⇔ \(x=2010\)
Vậy S = {2010}
PM
1
NM
Nguyễn Minh Quang
Giáo viên
18 tháng 3 2022
ta có :
\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)
hay x =2010
Vậy phương trình có nghiệm x = 2010
PB
1
BH
19 tháng 3 2019
\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
\(\Leftrightarrow x=2010\)
HT
2
QD
19 tháng 5 2018
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Rightarrow x=2010\)
Vậy....
21 tháng 7 2018
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(x-2010=0\)
\(x=2010\)
Vậy x = 2010
PT
2
20 tháng 3 2019
\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)
\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=2010\)
V...\(S=\left\{2010\right\}\)
^^
20 tháng 3 2019
\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\Leftrightarrow\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)
\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=2010\)
NN
0
B
23 tháng 3 2017
x-1009/1001+x-4/1003+x+2010/1005=7
((x-1009/1001)-1))+((x-4/1003)-2)+((x+2010/1005)-4))=0
(x-2010/1001)+(x-2010/1003)+(x-2010/1005)=0
(x-2010)*(1/1001+1/1003+1/1005)=0
okk!!!!!!!!!!!!!!!
a)\(\frac{x-10}{2010}\)+ \(\frac{x-3}{2003}\)+\(\frac{x-2}{2002}\)= -3
=> \(\frac{x-10}{2010}\)+1+ \(\frac{x-3}{2003}\)+ 1+\(\frac{x-2}{2002}\)+1= -3 +1 + 1 + 1
=> \(\frac{x-10+2010}{2010}\)+ \(\frac{x-3+2003}{2003}\)+\(\frac{x-2+2002}{2002}\)= 0
=>\(\frac{x+2000}{2010}\)+ \(\frac{x+2000}{2003}\)+\(\frac{x+2000}{2002}\)= 0
=>(x + 2000)(\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)) = 0
=> x + 2000 = 0
hoặc
=>\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)= 0
Mà : \(\frac{1}{2010}\)> 0
\(\frac{1}{2003}\)> 0
\(\frac{1}{2002}\)> 0
Cộng vế theo vế của các bất đẳng thức trên , ta có:
\(\frac{1}{2010}\)+\(\frac{1}{2003}\)+\(\frac{1}{2002}\)>0
=> x + 2000 = 0
=> x = 0 -2000 = -2000
Vậy x = -2000
Nhường các bạn câu 2 :(