Lời giải:
ĐK: \(x\geq -1\)
\(2(x^2+2)=5\sqrt{x^3+1}\)
\(\Leftrightarrow 2(x^2+2)=5\sqrt{(x+1)(x^2-x+1)}\)
Đặt \(\sqrt{x+1}=a; \sqrt{x^2-x+1}=b(a,b\geq 0)\)
\(\Rightarrow a^2+b^2=x^2+2\)
Khi đó pt trở thành:
\(2(a^2+b^2)=5ab\)
\(\Leftrightarrow 2(a^2+b^2)-5ab=0\)
\(\Leftrightarrow (2a-b)(a-2b)=0\Rightarrow \left[\begin{matrix} 2a=b\\ a=2b\end{matrix}\right.\)
Nếu \(2a=b\Rightarrow 4a^2=b^2\Rightarrow 4(x+1)=x^2-x+1\)
\(\Rightarrow x^2-5x-3=0\Rightarrow x=\frac{5\pm \sqrt{37}}{2}\) (t/m)
Nếu \(a=2b\Rightarrow a^2=4b^2\Rightarrow x+1=4(x^2-x+1)\)
\(\Rightarrow 4x^2-5x+3=0\Leftrightarrow (2x-\frac{5}{4})^2+\frac{23}{16}=0\) (vô nghiệm)
Vậy pt có nghiệm \(x=\frac{5\pm \sqrt{37}}{2}\)
