Question

решить неравенства
а) 3x^2+4x<16x(x+1)

Answer 1

$3x^2+4x<16x(x+1)\3x^2+4x<16x^2+16x\13x^2+12x>0\13x^2+12x=0\x=0 x=-frac{12}{13}\\////+///(-frac{12}{13})__-__(0)////+///>x\xin(-infty;-frac{12}{13})cup(0;+infty)$