Đặt x2=a(a≥0)Ta có pt⟺10√a3+1=3(a2+2)
⟺10(√a3+1−3a−3)=3a2+6−30a−30
⟺10(a3+1−9a2−18a−9)√a3+1+3a+3=3a2−30a−24
⟺10(a3−9a2−18a−8)√a3+1+3a+3=3(a2−10a−8)
⟺10(a2−10a−8)(a+1)√a3+1+3a+3=3(a2−10a−8)
⟺(a2−10a−8)(10(a+1)√a3+1+3a+3−3)=0
Ta có: a≥0=>10(a+1)√a3+1+3a+3−3<0 (BD tương đương)
Vậy a2−10a−8=0⟺a=5+√33(doa≥0)=>x=√5+√33,x=−√5+√33