Thế `m=2` vào (1) \(\Leftrightarrow x^2-7x+12=0\)
\(\Delta=\left(-7\right)^2-4.1.12=1>0\)
`->` ptr có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x=\dfrac{7+\sqrt{1}}{1}=4\\x=\dfrac{7-\sqrt{1}}{1}=3\end{matrix}\right.\)
Vậy \(S=\left\{3;4\right\}\)
b. \(\Delta=\left(-7\right)^2-4\left(2m+8\right)=49-8m-32=17-8m\)
Để ptr có 2 nghiệm \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow m\le\dfrac{17}{8}\)
Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=7\\x_1x_2=2m+8\end{matrix}\right.\)
Ta có: \(x_1^2+x_2^2=\left(x_1x_2-7\right)^2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=\left(x_1x_2-7\right)^2\)
\(\Leftrightarrow7^2-2\left(2m+8\right)=\left(2m+8-7\right)^2\)
\(\Leftrightarrow49-4m-16=4m^2+4m+1\)
\(\Leftrightarrow4m^2=32\)
\(\Leftrightarrow m^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}m=2\sqrt{2}\left(l\right)\\m=-2\sqrt{2}\left(n\right)\end{matrix}\right.\)
Vậy \(m=-2\sqrt{2}\) thỏa đề bài
