Giải:
ĐK: $x$ \geq $-1$
$2(x^2+2)=5\sqrt{x^3+1}$
$\iff 2[(x^2-x+1)+(x+1)]=5\sqrt{(x+1)(x^2-x+1)}$
Đặt $\sqrt{x+1}=a; \sqrt{x^2-x+1}=b ;(a,b$ \geq $0)$. Thay vào ta có:
Pt $\iff 2(a^2+b^2)-5ab=0$
$\iff (2a-b)(a-2b)=0$
$\iff 2a=b$ v $a=2b$
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