\(a,\text{Sửa: }x^2+5x-3=0\\ \Delta=25+12=37\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5-\sqrt{37}}{2}\\x=\dfrac{-5+\sqrt{37}}{2}\end{matrix}\right.\)
\(b,\Delta=\left(2\sqrt{3}-1\right)^2-4\left(4\sqrt{3}-6\right)=37-20\sqrt{3}=\left(5-2\sqrt{3}\right)^2>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}-1-5+2\sqrt{3}}{2}=-3+2\sqrt{3}\\x=\dfrac{2\sqrt{3}-1+5-2\sqrt{3}}{2}=2\end{matrix}\right.\)
\(c,\Delta'=\left(2\sqrt{3}\right)^2+4=4+12=16\\ \Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{3}-4\\x=2\sqrt{3}+4\end{matrix}\right.\)
b) \(x^2-\left(2\sqrt{3}-1\right)x+4\sqrt{3}-6=0\)
<=> \(\left(x-2\right)\left(x+2\right)-\left(2\sqrt{3}-1\right)\left(x-2\right)=0\)
<=> \(\left(x-2\right)\left(x-2\sqrt{3}+3\right)=0\)
<=> \(\left[{}\begin{matrix}x=2\\x=2\sqrt{3}-3\end{matrix}\right.\)
d) \(x^2-4\sqrt{3}x-4=0\)
<=> \(\left(x-2\sqrt{3}\right)^2=16\)
<=> \(\left[{}\begin{matrix}x=\left(\sqrt{3}+1\right)^2\\x=-\left(\sqrt{3}-1\right)^2\end{matrix}\right.\)
