\(co;2N_0\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\\Delta_x=\left(m+3\right)^2-4m\left(2m+1\right)=-7m^2+2m+9\ge0\end{matrix}\right.\)
\(\Delta'_m=1+63=64\)\(\rightarrow\Delta_x\ge0\Leftrightarrow-1\le m\le\dfrac{9}{7}\)
\(\Rightarrow m\in[-1;0)U(0;\dfrac{9}{7}]\)
với điều m có hệ thức vi-et: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{m+3}{m}\\x_1x_2=\dfrac{2m+1}{m}\end{matrix}\right.\)(I)
đầu bai : \(\left|x_1-x_2\right|=2\) \(\Leftrightarrow\left(x_1-x_2\right)^2=4\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4\) (a)
từ (I) và (a) \(\Leftrightarrow\left(\dfrac{m+3}{m}\right)^2-4\left(\dfrac{2m+1}{m}\right)=4\)
đặt 1/m =y
\(\Leftrightarrow\left(1+3y\right)^2-4\left(2+y\right)=4\)
đặt 1+3y =z
\(3z^2-4z-32=0\)
\(\Delta_z'=4+3.32=4\left(1+24\right)=\left(2.5\right)^2\)
\(\left[{}\begin{matrix}z_1=\dfrac{2-10}{3}=-\dfrac{8}{3};y_1=\dfrac{z_1-1}{3}=\dfrac{-\dfrac{8}{3}-1}{3}=-\dfrac{11}{9}\\z_2=\dfrac{2+10}{3}=4;y_2=\dfrac{z_2-1}{3}=\dfrac{4-1}{3}=1\end{matrix}\right.\)
\(m=\dfrac{1}{y};\rightarrow\left[{}\begin{matrix}m_1=\dfrac{1}{y_1}=-\dfrac{9}{11}\\m_2=\dfrac{1}{y_2}=\dfrac{1}{1}\end{matrix}\right.\) nhận hết
đạt giá trị nhỏ nhất
