Question

[Hồ thị quỳnh hương]

 File: 17136207288062370288475005508247.jpg 

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\text{Δ}=\left(m+3\right)^2-4\left(m-1\right)\)

\(=m^2+6m+9-4m+4\)

\(=m^2+2m+13=\left(m+1\right)^2+12>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+3\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(x_1< \dfrac{1}{2}< x_2\)

=>\(\left(x_1-\dfrac{1}{2}\right)\left(x_2-\dfrac{1}{2}\right)< 0\)

=>\(x_1x_2-\dfrac{1}{2}\left(x_1+x_2\right)+\dfrac{1}{4}< 0\)

=>\(m-1-\dfrac{1}{2}\left(m+3\right)+\dfrac{1}{4}< 0\)

=>\(m-\dfrac{3}{4}-\dfrac{1}{2}m-\dfrac{3}{2}< 0\)

=>\(\dfrac{1}{2}m< \dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)

=>\(m< \dfrac{9}{4}\cdot2=\dfrac{9}{2}\)