Để pt có 2 nghiệm trái dấu \(\Leftrightarrow2m-3< 0\Rightarrow m< \frac{3}{2}\)
b/ \(\Delta'=\left(m+1\right)^2-2m+3=m^2+4>0\)
Phương trình luôn có 2 nghiệm phân biệt \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=2m-3\end{matrix}\right.\)
\(A=M^2=\frac{\left(x_1+x_2\right)^2}{\left(x_1-x_2\right)^2}=\frac{\left(x_1+x_2\right)^2}{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(A=\frac{\left(2m+2\right)^2}{\left(2m+2\right)^2-4\left(2m-3\right)}=\frac{4m^2+8m+4}{4m^2+16}=\frac{m^2+2m+1}{m^2+4}\)
\(\Leftrightarrow Am^2+4A=m^2+2m+1\)
\(\Leftrightarrow\left(A-1\right)m^2-2m+4A-1=0\)
\(\Delta'=1-\left(A-1\right)\left(4A-1\right)\ge0\)
\(\Leftrightarrow-4A^2+5A\ge0\)
\(\Leftrightarrow0\le A\le\frac{5}{4}\)
\(\Rightarrow A_{max}=\frac{5}{4}\) khi \(m=4\) hay \(M_{max}=\frac{\sqrt{5}}{2}\) khi \(m=4\)
