Với \(-\sqrt{3}< x< \sqrt{3}\) ta có :
\(\left(x^2-3\right)< 0;\left(x-\sqrt{3}\right)< 0\)
\(\Rightarrow3-x^2=\sqrt{3}-x\)
\(\Leftrightarrow\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)=\left(\sqrt{3}-x\right)\)
\(\Leftrightarrow\left(\sqrt{3}-x\right)\left(\sqrt{3}-1+x\right)=0\)
\(\left[{}\begin{matrix}x=\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)(t/m)
Với \(x\le-\sqrt{3}\)
\(\Rightarrow\left(x^2-3\right)\ge0;\left(x-\sqrt{3}\right)< 0\)
\(\Rightarrow x^2-3=\sqrt{3}-x\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\left(l\right)\\x=-\sqrt{3}-1\left(tm\right)\end{matrix}\right.\)
Với \(x\ge\sqrt{3}\)
\(\Rightarrow\left(x^2-3\right)\ge0;\left(x-\sqrt{3}\right)\ge0\)
\(\Rightarrow\left(x^2-3\right)=\left(x-\sqrt{3}\right)\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\left(tm\right)\\x=1-\sqrt{3}\left(l\right)\end{matrix}\right.\)
Vậy Phương trình đã cho có tập nghiệm \(S=\left\{\sqrt{3};1-\sqrt{3};-\sqrt{3}-1\right\}\)
