B=(998-1000)+(1002-1004)+.....+(2010-2012)

  =(-2)+(-2)+.............+(-2)

  =(-2).254

 =-508

e ngu toán

giúp e

1004 kg

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}

nếu a>0 ;b<0

\(\Rightarrow\left\{{}\begin{matrix}3a^2b^{1004}>0\\-19a^5b^{1008}< 0\end{matrix}\right.\)(ko thỏa mãn đề bài)

nếu a<0;b>0

\(\Rightarrow\left\{{}\begin{matrix}3a^2b^{1004}>0\\-19a^5b^{1008}>0\end{matrix}\right.\)(thỏa mãn)

vậy a là dấu âm,b là dấu dương

\(\dfrac{1004}{1005}< \dfrac{1005}{1006}< \dfrac{1006}{1007}< \dfrac{1007}{1008}\)

1004/1005<1005/1006<1006/1007<1007/1008

=1004.(1004+1):2= 1009020 : :2=504150

B.1006+1010 bạn nhé