a: \(\text{Δ}=\left[2\left(k+3\right)\right]^2-4\left(-k-1\right)\)
\(=\left(2k+6\right)^2+4\left(k+1\right)\)
\(=4k^2+24k+36+4k+4\)
\(=4k^2+28k+40\)
\(=\left(2k\right)^2+2\cdot2k\cdot7+49-9\)
\(=\left(2k+7\right)^2-9+=\left(2k+4\right)\left(2k+10\right)\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\left(k+3\right)\\x_1x_2=\dfrac{c}{a}=-k-1\end{matrix}\right.\)
Để phương trình có hai nghiệm dương thì \(\left\{{}\begin{matrix}\text{Δ}>0\\x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2k+4\right)\left(2k+10\right)>0\\-2\left(k+3\right)>0\\-k-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(k+2\right)\left(k+5\right)>0\\k+3< 0\\k+1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}k>-2\\k< -5\end{matrix}\right.\\k< -3\end{matrix}\right.\)
=>k<-5
b: \(x_1+x_2-2x_1x_2=-2\left(k+3\right)-2\cdot\left(-k-1\right)\)
\(=-2k-6+2k+2=-4\)
=>Đây là hệ thức không phụ thuộc vào tham số k
