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Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)
\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)
mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)
nên x+2013=0
hay x=-2013
Vậy: S={-2013}
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)
\(\Leftrightarrow x-2007=0\)
\(\Leftrightarrow x=2007\)
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
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Rightarrow x=2010\)
Vậy....
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(x-2010=0\)
\(x=2010\)
Vậy x = 2010
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Rightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Rightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}\right)=0\)
Dễ thấy: \(\dfrac{1}{1000}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}>0\Leftrightarrow x-2007=0\Leftrightarrow x=2007\)
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97
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
\(a.\Leftrightarrow\frac{3\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-9}{\left(x+1\right)\left(x-2\right)}.DKXD:x\ne-1;x\ne2\)
\(\Rightarrow3x-6-x-1=-9\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\)
\(b.\frac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.DKXDx\ne1;-1\)
\(\Rightarrow x^2+x-4x-4+x^2-x+4x-4=2x^2+2x-2x-2\)
\(\Leftrightarrow-6=0\left(voly\right)\)
vay \(S=\varnothing\)
\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
thks