Câu hỏi của Nguyễn Chí Thành

Question

Giải pt: $\sqrt{x^4+x^2+1}+\sqrt{3}\left(x^2+1\right)=3\sqrt{3}x$

Nguyễn Việt Lâm · Accepted Answer

$\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}+\sqrt{3}\left(x^2-3x+1\right)=0$

Đặt $\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\sqrt{x^2+x+1}=b>0\end{matrix}\right.$

$\Rightarrow ab+\sqrt{3}\left(2a^2-b^2\right)=0$

$\Leftrightarrow\left(\sqrt{3}a+b\right)\left(2a-\sqrt{3}b\right)=0$

$\Leftrightarrow4a^2=3b^2$

$\Leftrightarrow4\left(x^2-x+1\right)=3\left(x^2+x+1\right)$

$\Leftrightarrow...$Câu trả lời: $\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}+\sqrt{3}\left(x^2-3x+1\right)=0$

Đặt $\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\sqrt{x^2+x+1}=b>0\end{matrix}\right.$

$\Rightarrow ab+\sqrt{3}\left(2a^2-b^2\right)=0$

$\Leftrightarrow\left(\sqrt{3}a+b\right)\left(2a-\sqrt{3}b\right)=0$

$\Leftrightarrow4a^2=3b^2$

$\Leftrightarrow4\left(x^2-x+1\right)=3\left(x^2+x+1\right)$

$\Leftrightarrow...$

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