Lời giải:
\(4(x^2+4x+2)=11\sqrt{x^4+4}=11\sqrt{(x^2+2)^2-4x^2}\)
\(\Leftrightarrow 4(x^2+4x+2)=11\sqrt{(x^2+2+2x)(x^2+2-2x)}\)
\(\Leftrightarrow 6(x^2+2+2x)-2(x^2+2-2x)=11\sqrt{(x^2+2+2x)(x^2+2-2x)}\)
Đặt \(\sqrt{x^2+2+2x}=a; \sqrt{x^2+2-2x}=b(a,b\geq 0)\)
Khi đó pt đã cho trở thành:
\(6a^2-2b^2=11ab\)
\(\Leftrightarrow 6a^2-11ab-2b^2=0\)
\(\Leftrightarrow 6a(a-2b)+b(a-2b)=0\)
\(\Leftrightarrow (6a+b)(a-2b)=0\)
Nếu \(6a+b=0\). Vì \(a,b\geq 0\) nên để \(6a+b=0\) thì \(a=b=0\)
\(\Rightarrow \sqrt{x^2+2+2x}=\sqrt{x^2+2-2x}=0\) (vô lý)
Nếu \(a-2b=0\Leftrightarrow a=2b\)
\(\Rightarrow a^2=4b^2\Leftrightarrow x^2+2+2x=4(x^2+2-2x)\)
\(\Leftrightarrow 3x^2-10x+6=0\Rightarrow x=\frac{5\pm \sqrt{7}}{3}\) (t.m)
Vậy \(x=\frac{5\pm \sqrt{7}}{2}\)
