Question

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

Answer 1

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