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Đặt $x^2=a(a\ge 0)$Ta có $pt\iff 10\sqrt{a^3+1}=3(a^2+2)$$\iff 10(\sqrt{a^3+1}-3a-3)=3a^2+6-30a-30$$\iff \frac{10(a^3+1-9a^2-18a-9)}{\sqrt{a^3+1}+3a+3}=3a^2-30a-24$$\iff \frac{10(a^3-9a^2-18a-8)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff \frac{10(a^2-10a-8)(a+1)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff (a^2-10a-8)(\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3)=0$Ta có: $a\ge 0=> \sqrt{a^3+1}+3a+3>3a+3=3(a+1)=>\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3<\frac{10}{3}-3&lt;0$Vậy $a^2-10a-8=0\iff a=5+\sqrt{33}(do a\ge 0)=>x=\sqrt{5+\sqrt{33}},x=-\sqrt{5+\sqrt{33}} $

Đặt $x^2=a(a\ge 0)$Ta có $pt\iff 10\sqrt{a^3+1}=3(a^2+2)$$\iff 10(\sqrt{a^3+1}-3a-3)=3a^2+6-30a-30$$\iff \frac{10(a^3+1-9a^2-18a-9)}{\sqrt{a^3+1}+3a+3}=3a^2-30a-24$$\iff \frac{10(a^3-9a^2-18a-8)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff \frac{10(a^2-10a-8)(a+1)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff (a^2-10a-8)(\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3)=0$Ta có: $a\ge 0=>\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3<0$ (BD tương đương)Vậy $a^2-10a-8=0\iff a=5+\sqrt{33}(do a\ge 0)=>x=\sqrt{5+\sqrt{33}},x=-\sqrt{5+\sqrt{33}} $

Đặt $x^2=a(a\ge 0)$Ta có $pt\iff 10\sqrt{a^3+1}=3(a^2+2)$$\iff 10(\sqrt{a^3+1}-3a-3)=3a^2+6-30a-30$$\iff \frac{10(a^3+1-9a^2-18a-9)}{\sqrt{a^3+1}+3a+3}=3a^2-30a-24$$\iff \frac{10(a^3-9a^2-18a-8)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff \frac{10(a^2-10a-8)(a+1)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff (a^2-10a-8)(\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3)=0$Ta có: $a\ge 0=> \sqrt{a^3+1}+3a+3>3a+3=3(a+1)=>\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3<\frac{10}{3}-3<0$Vậy $a^2-10a-8=0\iff a=5+\sqrt{33}(do a\ge 0)=>x=\sqrt{5+\sqrt{33}} $

Đặt $x^2=a(a\ge 0)$Ta có $pt\iff 10\sqrt{a^3+1}=3(a^2+2)$$\iff 10(\sqrt{a^3+1}-3a-3)=3a^2+6-30a-30$$\iff \frac{10(a^3+1-9a^2-18a-9)}{\sqrt{a^3+1}+3a+3}=3a^2-30a-24$$\iff \frac{10(a^3-9a^2-18a-8)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff \frac{10(a^2-10a-8)(a+1)}{\sqrt{a^3+1}+3a+3}=3(a^2-10a-8)$$\iff (a^2-10a-8)(\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3)=0$Ta có: $a\ge 0=> \sqrt{a^3+1}+3a+3>3a+3=3(a+1)=>\frac{10(a+1)}{\sqrt{a^3+1}+3a+3}-3<\frac{10}{3}-3<0$Vậy $a^2-10a-8=0\iff a=5+\sqrt{33}(do a\ge 0)=>x=\sqrt{5+\sqrt{33}},x=-\sqrt{5+\sqrt{33}} $