\(a,\) Thay \(m=5\) vào pt trên :
\(\Rightarrow x^2+5x+\left(-5\right)^2+3.5-1=0\)
\(\Rightarrow x^2+5x+25+15-1=0\)
\(\Rightarrow x^2+5x+5=0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{5}}{2}\\x_2=\dfrac{-5-\sqrt{5}}{2}\end{matrix}\right.\)
\(b,\) Thay \(x=2\) vào pt trên :
\(\Rightarrow2^2+2m-m^2+3m-1=0\)
\(\Rightarrow-m^2+5m+3=0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{5+\sqrt{37}}{2}\\x_2=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\)
\(c,\)Để pt có nghiệm kép thì \(\Delta=0\)
\(\Delta=b^2-4ac=m^2-4\left(-m^2+3m-1\right)=m^2+4m^2-12m+4=5m^2-12m+4\)
\(\Rightarrow5m^2-12m+4=0\)
\(\Rightarrow\left\{{}\begin{matrix}m_1=2\\m_2=\dfrac{2}{5}\end{matrix}\right.\)
