Bài 17 :
a) \(x^2-\left(2m+1\right)x+m^2+m-6=0\left(1\right)\)
\(\Delta=\left(2m+1\right)^2-4\left(m^2+m-6\right)\)
\(\Delta=4m^2+4m+1-4m^2-4m+24\)
\(\Delta=25>0\)
\(Pt\left(1\right)\) luôn có 2 nghiệm phân biệt
b) Để \(\left(1\right)\) có 2 nghiệm âm phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S< 0\\P>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25>0,\forall m\\2m+1< 0\\m^2+m-6>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< -\dfrac{1}{2}\\\left[{}\begin{matrix}m< -3\\m>2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow m< -3\)
Vậy \(m< -3\) thỏa mãn đề bài
c) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-x_1x_2\)
\(\Rightarrow A=S^2-2P=\left(2m+1\right)^2-2\left(m^2+m-6\right)\)
\(\Rightarrow A=4m^2+4m+1-2m^2-2m+12\)
\(\Rightarrow A=2m^2+2m+13\)
\(\Rightarrow A=2\left(m^2+m+\dfrac{1}{4}\right)+13-\dfrac{1}{2}\)
\(\)\(\Rightarrow A=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{25}{2}\ge\dfrac{25}{2}\) \(\left(2\left(m+\dfrac{1}{2}\right)^2\ge0,\forall m\in R\right)\)
\(\Rightarrow GTNN\left(A\right)=\dfrac{25}{2}\left(tại.x=-\dfrac{1}{2}\right)\)
Bài 12 :
a) \(16x^2-17x+1=0\left(1\right)\)
Ta thấy : \(a+b+c=16-17+1=0\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{16}\end{matrix}\right.\)
b) \(2x^2-4x-6=0\left(1\right)\)
Ta thấy : \(a-b+c=2+4-6=0\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
c) \(2x^2-40x+38=0\left(1\right)\)
Ta thấy : \(a+b+c=2-40+38=0\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\x=19\end{matrix}\right.\)
d) \(1230x^2-5x-1235=0\left(1\right)\)
Ta thấy : \(a-b+c=1230+5-1235=0\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1235}{1230}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{247}{246}\end{matrix}\right.\)
