Bài 10:
a: DKXĐ: x<>1
\(Q=\dfrac{3+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-1}{x^2+x+1}\)
b: x^2=1
=>x=1(loại) hoặc x=-1(nhận)
Khi x=-1 thì \(Q=\dfrac{-1}{\left(-1\right)^2+\left(-1\right)+1}=\dfrac{-1}{1}=-1\)
c: x^2+x+1=(x+1/2)^2+3/4>0
=>Q<0
