Question

Tìm x để \(\dfrac{4x^2-8x}{-x^2+x+6}< 0\)

Answer 1

ĐKXĐ: x∉{3;-2}

Ta có: \(\frac{4x^2-8x}{-x^2+x+6}<0\)

=>\(\frac{4x\left(x-2\right)}{-\left(x^2-x-6\right)}<0\)

=>\(\frac{x\left(x-2\right)}{x^2-x-6}>0\)

=>\(\frac{x\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}>0\)

TH1: \(\begin{cases}x\left(x-2\right)>0\\ \left(x-3\right)\left(x+2\right)>0\end{cases}\)

=>\(\begin{cases}\left[\begin{array}{l}x>2\\ x<0\end{array}\right.\\ \left[\begin{array}{l}x>3\\ x<-2\end{array}\right.\end{cases}\)

=>x>3 hoặc x<-2

TH2: \(\begin{cases}x\left(x-2\right)<0\\ \left(x-3\right)\left(x+2\right)<0\end{cases}\)

=>\(\begin{cases}0

=>0<x<2