Câu 1: \(x^2-mx+\left(m-2\right)^2=0\)
Ta có: \(\Delta=-3m^2+16m-16\)
Để phương trình có 2 nghiệm \(\Leftrightarrow\Delta\ge0\) \(\Leftrightarrow\frac{4}{3}\le m\le4\)
Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1\cdot x_2=m^2-4m+4\end{matrix}\right.\)
Mặt khác: \(A=x_1x_2+2\left(x_1+x_2\right)\)
\(\Rightarrow A=m+2m^2-8m+8=2m^2-7m+8\) \(=2\left(m^2-\frac{7}{2}m+4\right)=2\left(m^2-2\cdot m\cdot\frac{7}{4}+\frac{49}{16}+\frac{15}{16}\right)\) \(=2\left(m-\frac{7}{4}\right)^2+\frac{15}{8}\ge\frac{15}{8}\)
Dấu bằng xảy ra \(\Leftrightarrow m-\frac{7}{4}=0\) \(\Leftrightarrow m=\frac{7}{4}\left(TM\right)\)
Vậy \(Min_A=\frac{15}{8}\) khi \(m=\frac{7}{4}\)
Câu 2: \(mx^2+2\left(m-2\right)x+m-3=0\)
Ta có: \(\Delta'=4-m\)
Để phương trình có 2 nghiệm \(\Leftrightarrow m\le4\)
Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\frac{4-2m}{m}\\x_1\cdot x_2=\frac{m-3}{m}\end{matrix}\right.\) \(\left(m\ne0\right)\)
Mặt khác: \(\frac{x_1}{x_2}+\frac{x_2}{x_1}=3\) \(\Leftrightarrow\frac{x_1^2+x_2^2}{x_1x_2}=3\) \(\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=3\) \(\left(m\ne3\right)\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2=0\)
\(\Rightarrow\frac{16-16m+4m^2}{m^2}-\frac{5m^2-15m}{m^2}=0\)
\(\Leftrightarrow\frac{16-m-m^2}{m}=0\) \(\Leftrightarrow16-m-m^2=0\) \(\Leftrightarrow m=\frac{-1\pm\sqrt{65}}{2}\left(TM\right)\)
Vậy \(m=\frac{-1\pm\sqrt{65}}{2}\)
