\(x^2+mx+m+3=0\)
\(\Delta=m^2-4\cdot\left(m+3\right)\)
\(=m^2-4m-12\)
\(=\left(m-6\right)\left(m+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\le-2\\m\ge6\end{matrix}\right.\)
Theo định lý Viet ta có :
\(\left\{{}\begin{matrix}x_1+x_2=\frac{-m}{2}\\x_1x_2=\frac{m+3}{2}\end{matrix}\right.\)
Từ đó ta có hệ :
\(\left\{{}\begin{matrix}x_1+x_2=\frac{-m}{2}\\x_1x_2=\frac{m+3}{2}\\2x_1+3x_2=5\end{matrix}\right.\)
Pt cuối \(\Leftrightarrow2\left(x_1+x_2\right)+x_2=5\)
\(\Leftrightarrow-m+x_2=5\)
\(\Leftrightarrow x_2=m+5\)(1)
Thay lên pt đầu: \(m+5+x_1=\frac{-m}{2}\)
\(\Leftrightarrow x_1=\frac{-m}{2}-\frac{2\left(m+5\right)}{2}\)
\(\Leftrightarrow x_1=\frac{-m-2m-10}{2}=\frac{-3m-10}{2}\)(2)
Thay (1) và (2) vào pt giữa :
\(\left(m+5\right)\cdot\frac{-3m-10}{2}=\frac{m+3}{2}\)
\(\Leftrightarrow m=\frac{-13\pm\sqrt{10}}{3}\)( thỏa )
Vậy...
Is that true .-.
