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