a) Ta có: \(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(m-3\right)=4-4m+12=-4m+16\)
Để phương trình có hai nghiệm thì \(\text{Δ}\ge0\)
\(\Leftrightarrow-4m+16\ge0\)
\(\Leftrightarrow-4m\ge-16\)
hay \(m\le4\)
Ta có: \(\Delta'=4-m\)
a) Để phương trình có 2 nghiệm \(\Leftrightarrow\Delta'\ge0\) \(\Leftrightarrow m\le4\)
b) Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-3\end{matrix}\right.\)
Ta có: \(x_1^3+x_2^3=-20\)
\(\Rightarrow\left(x_1+x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)=-20\)
\(\Rightarrow\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=-20\)
\(\Rightarrow2\left[2^2-3\left(m-3\right)\right]=-20\)
\(\Leftrightarrow8-6m+18=-20\) \(\Leftrightarrow m=\dfrac{23}{3}\) (Thỏa mãn)
Vậy \(m=\dfrac{23}{3}\)
