\(\left(x-3\right)\left(x-5\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3< 0\\x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3>0\\x-5< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x>5\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x< 5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow3< x< 5\)
câu 2
a. \(\left\{{}\begin{matrix}-7\le x< 7\\x\in Z\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6\right\}\)
Gọi S là tổng các số nguyên x thỏa\(\left\{{}\begin{matrix}-7\le x< 7\\x\in Z\end{matrix}\right.\)
S= (-7)+(-6)+(-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6
=(-7)+ (-6)+6+(-5)+5+(-4)+4+(-3)+3+(-2)+2+(-1)+1+0
= -7
câu 2.b bạn làm tương tự
\(\left|x\right|< 2013\Leftrightarrow-2013< x< 2013\)
\(\Leftrightarrow x\in\left\{-2012,-2011,-2010,....,-2,-1,0,1,2,....,2010,2011,2012\right\}\)
gọi S là tổng cần tìm
=> S = (-2012)+(-2011)+....+(-1)+0+1+...+2011+2012
=> S = (-2012)+2012+(-2011)+2011+....+(-1)+1+0
=>S=0
