bài 1:
a: \(\dfrac{x+3}{x-7}>0\)
TH1: \(\left\{{}\begin{matrix}x+3>0\\x-7>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-3\\x>7\end{matrix}\right.\)
=>x>7
TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-7< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -3\\x< 7\end{matrix}\right.\)
=>x<-3
Vậy: \(x\in Z\backslash\left\{-3;-2;....;6;7\right\}\)
b: \(\dfrac{x-5}{x-10}< 0\)
TH1: \(\left\{{}\begin{matrix}x-5>0\\x-10< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>5\\x< 10\end{matrix}\right.\)
=>5<x<10
mà x nguyên
nên \(x\in\left\{6;7;8;9\right\}\)
2)
a) Ta có:
`-1/4<0`
`1/100>0`
`=>-1/4<1/100`
b) Ta có:
`2<3=>1/2>1/3`
`=>-1/2<-1/3`
c) Ta có:
`-2/3<0`
`-3/-5=3/5>0`
`=>-2/3<3/5`
d) Ta có:
`-2,5=(-2,5*2)/2=-5/2=5/-2`