\(\sqrt{x^4+3x^2-4}+3x=\sqrt{3x^4+16}\)
\(\Leftrightarrow\sqrt{3x^4+16}-\sqrt{x^4+3x^2-4}=3x\)
\(\Leftrightarrow4x^4+3x^2+12-2\sqrt{3x^4+16}.\sqrt{x^4+3x^2-4}=9x^2\)
Đặt \(x^2=a\ge0\)
\(\Leftrightarrow2a^2-3a+6=\sqrt{3a^2+16}.\sqrt{a^2+3a-4}\)
\(\Leftrightarrow\left(2a^2-3a+6\right)^2=\left(3a^2+16\right).\left(a^2+3a-4\right)\)
\(\Leftrightarrow\left(a^2+4\right)\left(a^2-21a^2+25\right)=0\)
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