\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)

\(\Rightarrow(\frac{x-1009}{1010}-1)+\left(\frac{x-1007}{1012}-1\right)=\left(\frac{x-1010}{1009}-1\right)+\left(\frac{x-1012}{1007}-1\right)\)

\(\Rightarrow\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}\)

\(\Rightarrow\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)

Ta có

\(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\ne0\Rightarrow x-2019=0\Rightarrow x=2019\)

\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)

\(\frac{x-1009}{1010}-1+\frac{x-1007}{1012}-1=\frac{x-1010}{1009}-1+\frac{x-1012}{1007}\)\(\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}=0\)

\(\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)

1/1010 + 1/1012 - 1/1009 - 1/1007 khác 0

=> x - 2019 =0 => x = 2019

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}

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

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

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

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

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

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

\(\Rightarrow x+2013=0\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

\(B=\frac{1010+1007+\frac{2017}{113}+\frac{2017}{117}-\frac{1010}{119}-\frac{1007}{119}}{1010+1008+\frac{2018}{113}+\frac{2018}{117}-\frac{1010}{119}-\frac{1008}{119}}\)

\(B=\frac{2017+\frac{2017}{113}+\frac{2017}{117}-\frac{2017}{119}}{2018+\frac{2018}{113}+\frac{2018}{117}-\frac{2018}{119}}\)

\(B=\frac{2017.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}{2018.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)

\(B=\frac{2017}{2018}\)

Vậy \(B=\frac{2017}{2018}\)

Chúc bạn học tốt !!! 

dễ lắm thử nghỉ đi banj

Ta có : Q=\(\frac{1010+1011+1012}{1011+1012+1013}\)=\(\frac{1010}{1011+1012+1013}+\frac{1011}{1011+1012+1013}+\frac{1012}{1011+1012+1013}\)

Vì1010/1011>1010/1011+1012+1013

    1011/1012>1011/1011+1012+1013

    1012/1013>1012/1011+1012+1013

    =>P>Q

a, \(\frac{x+1006}{1000}+\frac{x+1007}{999}+\frac{x+1008}{998}+\frac{x+1009}{997}+\frac{x+2022}{4}=0\)

\(\Leftrightarrow\frac{x+1006}{1000}+1+\frac{x+1007}{999}+1+\frac{x+1008}{998}+1+\frac{x+1009}{997}+1+\frac{x+2022}{4}-4=0\)

\(\Leftrightarrow\frac{x+2006}{1000}+\frac{x+2006}{999}+\frac{x+2006}{998}+\frac{x+2006}{997}+\frac{x+2006}{4}=0\)

\(\Leftrightarrow\left(x+2006\right)\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\right)=0\)

Mà \(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\ne0\)

\(\Rightarrow x+2006=0\Leftrightarrow x=-2006\)