Question

решите неравенство ((3X^2+4x-4)/(8+15x))<0

Answer 1

$frac{3x^{2}+4x-4}{8+15x}<0\\ ODZ:\\ 8+15x neq 0\ 15x neq -8 | :15\ xneq-frac{8}{15}\\ 3x^{2}+4x-4 = 0\ D=b^{2}-4ac=4^{2}-4cdot 3cdot (-4)=16-(-48)=64\ x_{1,2}=frac{-b+/- sqrt{D}}{2a}\ x_{1}=frac{-4+8}{2cdot 3}=frac{4}{6}=frac{2}{3}\ x_{2}=frac{-4-8}{2cdot 3}=frac{-12}{6}=-2\\ frac{3(x-frac{2}{3})(x+2)}{8+15x}<0\ x<frac{2}{3}\\ x<-2$

 

/////////////-2////////////⅔------------------------------->x

 

x ∈ (-∞; -2) U (-2; ⅔)

 

Ответ: x ∈ (-∞; -2) U (-2; ⅔).