Lời giải:
a)
\(x^2+4x-5=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x(x-1)+5(x-1)=0\)
\(\Leftrightarrow (x+5)(x-1)=0\Rightarrow \left[\begin{matrix} x+5=0\\ x-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-5\\ x=1\end{matrix}\right.\)
b)
\(x^2-4x-1=0\)
\(\Leftrightarrow x^2-4x+4-5=0\)
\(\Leftrightarrow (x-2)^2-5=0\)
\(\Leftrightarrow (x-2-\sqrt{5})(x-2+\sqrt{5})=0\)
\(\Rightarrow \left[\begin{matrix} x-2-\sqrt{5}=0\\ x-2+\sqrt{5}=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2+\sqrt{5}\\ x=2-\sqrt{5}\end{matrix}\right.\)
c)
\(4x^2+24x+9=0\)
\(\Leftrightarrow (2x)^2+2.(2x).6+6^2-27=0\)
\(\Leftrightarrow (2x+6)^2-27=0\)
\(\Leftrightarrow (2x+6-\sqrt{27})(2x+6+\sqrt{27})=0\)
\(\Rightarrow \left[\begin{matrix} 2x+6-\sqrt{27}=0\\ 2x+6+\sqrt{27}=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-6+\sqrt{27}}{2}\\ x=\frac{-6-\sqrt{27}}{2}\end{matrix}\right.\)
