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25 tháng 3 2018

\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)

\(\Leftrightarrow x-2007=0\)

\(\Leftrightarrow x=2007\)

6 tháng 12 2017

\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Rightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Rightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}\right)=0\)

Dễ thấy: \(\dfrac{1}{1000}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}>0\Leftrightarrow x-2007=0\Leftrightarrow x=2007\)

3 tháng 3 2017

\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)

\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)

\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)

\(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)

\(\Rightarrow x+2013=0\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

3 tháng 3 2017

thks

NM
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

7 tháng 11 2019

Câu a : B-A= 10012 +10022 +10042 +10072 -10002-10032-10052-10062

=(10012-10002)+(10022-10032) (10042-10052)+(10072-10062)

HĐT số 3= (1001-1000)*(1001+1000)+(1002-1003)*(1002+1003)+(1004-1005)*(1004+1005)+(1007-1006)*(1007+1006)

=2001 -2005-2009+2013=0

vậy A=B

10 tháng 5 2017

Ta có:B-A=10012+10022+10042+10072-10002-10032-10052-10062

=(10012-1000)2+(10022-10032)+(10042-10052)+(10072-10062)

=(1001-1000)(1001+1000)+(1002-1003)(1002+1003)+(1004-1005)(1004+1005)+(1007-1006)(1007+1006)

=2001-2005-2009+2013

=0

=>A=B

10 tháng 5 2017

cảm ơn nha !!! thanghoa

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