Question

Answer 1

\(\Delta=\left(m-2\right)^2\ge0;\forall m\) nên pt luôn có 2 nghiệm

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)

\(A=\dfrac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\dfrac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}\)

\(=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}\)

\(A=\dfrac{m^2+2-m^2+2m-1}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)

\(A_{max}=1\) khi \(m=1\)

\(A=\dfrac{4m+2}{2\left(m^2+2\right)}=\dfrac{-\left(m^2+2\right)+m^2+4m+4}{2\left(m^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(m+2\right)^2}{2\left(m^2+2\right)}\ge-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(m=-2\)