ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+\sqrt{3}}=a>0\\\sqrt{x^2-\sqrt{3}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2x^2\)
\(\Rightarrow\dfrac{a^2}{x+a}+\dfrac{b^2}{x-b}=x\)
\(\Rightarrow a^2\left(x-b\right)+b^2\left(x+a\right)=x\left(x+a\right)\left(x-b\right)\)
\(\Leftrightarrow\left(a^2+b^2\right)x-a^2b+ab^2=x^3+ax^2-bx^2-abx\)
\(\Leftrightarrow2x^3-a^2b+ab^2=x^3+\left(a-b\right)x^2-abx\)
\(\Leftrightarrow x^3-\left(a-b\right)x^2+abx-a^2b+ab^2=0\)
\(\Leftrightarrow x^2\left(x-a+b\right)+ab\left(x-a+b\right)=0\)
\(\Leftrightarrow\left(x^2+ab\right)\left(x-a+b\right)=0\)
\(\Leftrightarrow x-a+b=0\)
\(\Leftrightarrow x+\sqrt{x^2-\sqrt{3}}=\sqrt{x^2+\sqrt{3}}\)
\(\Rightarrow2x^2-\sqrt{3}+2x\sqrt{x^2-\sqrt{3}}=x^2+\sqrt{3}\)
\(\Leftrightarrow2x\sqrt{x^2-\sqrt{3}}=2\sqrt{3}-x^2\)
\(\Rightarrow4x^2\left(x^2-\sqrt{3}\right)=12+x^4-4\sqrt{3}x^2\)
\(\Rightarrow x^4=4\)
\(\Rightarrow x=\pm\sqrt{2}\)
Thử lại chỉ có \(x=\sqrt{2}\) thỏa mãn
