\(\Delta'=4m^2-3m^2+3=m^2+3>0\)
Phương trình luôn có 2 nghiệm pb
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=4m\\x_1x_2=3m^2-3\end{matrix}\right.\)
\(A=\left|\frac{x_1+x_2+4}{x_1-x_2}\right|\Rightarrow A^2=\frac{\left(x_1+x_2+4\right)^2}{\left(x_1-x_2\right)^2}\)
\(B=A^2=\frac{\left(x_1+x_2+4\right)^2}{\left(x_1+x_2\right)^2-4x_1x_2}=\frac{\left(4m+4\right)^2}{4m^2+12}=\frac{4m^2+8m+4}{m^2+3}\)
\(\Leftrightarrow\left(B-4\right)m^2-8m+3B-4=0\)
\(\Delta'=16-\left(B-4\right)\left(3B-4\right)=-3B^2+16B\ge0\)
\(\Rightarrow0\le B\le\frac{16}{3}\) \(\Rightarrow B_{max}=\frac{16}{3}\) khi \(m=3\)
Hay \(A_{max}=\frac{4\sqrt{3}}{3}\) khi \(m=3\)
