Question

Помогите решить уравнение 9x^4-10x^2+1 >0

Answer 1

$9x^4-10x^2+1>0\x^2=t,t>0\9t^2-10t+1=0\D=64\t=cfrac{10pm 8}{18}\t=1\t=cfrac{1}{9}\x=pm 1\x=pmcfrac{1}{3}\(x+1)(x-1)left(x-cfrac{1}{3}right)left(x+cfrac{1}{3}right)>0\xin (-infty;-1)cup left(-cfrac{1}{3};cfrac{1}{3}right)cup (1;+infty)$