![image](https://tex.z-dn.net/?f=9x%5E4-10x%5E2%2B1%3E0%5C%5Cx%5E2%3Dt%2Ct%3E0%5C%5C9t%5E2-10t%2B1%3D0%5C%5CD%3D64%5C%5Ct%3D%5Ccfrac%7B10%5Cpm+8%7D%7B18%7D%5C%5Ct%3D1%5C%5Ct%3D%5Ccfrac%7B1%7D%7B9%7D%5C%5Cx%3D%5Cpm+1%5C%5Cx%3D%5Cpm%5Ccfrac%7B1%7D%7B3%7D%5C%5C%28x%2B1%29%28x-1%29%5Cleft%28x-%5Ccfrac%7B1%7D%7B3%7D%5Cright%29%5Cleft%28x%2B%5Ccfrac%7B1%7D%7B3%7D%5Cright%29%3E0%5C%5Cx%5Cin+%28-%5Cinfty%3B-1%29%5Ccup+%5Cleft%28-%5Ccfrac%7B1%7D%7B3%7D%3B%5Ccfrac%7B1%7D%7B3%7D%5Cright%29%5Ccup+%281%3B%2B%5Cinfty%29)
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)" alt="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)" align="absmiddle" class="latex-formula">