Question

cho phương trình x²-(m-1)x-5+m=0 (m là tham số)

tìm m để ( x1²-mx1+m-1)(x2²-mx2+m-1) < 0

Answer 1

bài này ko khó nhưng tính phê~~

xét pt đã cho : \(\Delta=\left(m-1\right)^2-4\left(m-5\right)=m^2-6m+21=\left(m-3\right)^2+12>12\)

=> pt luôn có 2 nghiệm pb với mọi m thuộc R

theo đl Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=m-5\end{matrix}\right.\)

\(\left(x_1^2-mx_1+m-1\right)\left(x_2^2-mx_2+m-1\right)< 0\)

\(\Leftrightarrow\left(x_1x_2\right)^2-mx_1x_2\left(x_1+x_2\right)+m\left(x_1+x_2\right)^2-2mx_1x_2-\left(x_1+x_2\right)^2+2x_1x_2+m^2x_1x_2-m^2\left(x_1+x_2\right)+m\left(x_1+x_2\right)+m^2-2m+1< 0\)

\(\Leftrightarrow\left(m-5\right)^2-m\left(m-5\right)\left(m-1\right)+m\left(m-1\right)^2-2m\left(m-5\right)-\left(m-1\right)^2+2\left(m-5\right)+m^2\left(m-5\right)-m^2\left(m-1\right)+m\left(m-1\right)+m^2-2m+1< 0\)

\(\Leftrightarrow-3m+15< 0\Leftrightarrow m>5\)

Kl: m > 5

Answer 2

x1 =a; x_2=b

<=> (a^2 -(m-1)a +m-1 -a -5)(a^2 -(m-1)b +m-1 -b -5) <0

<=> (a +5)(b+5) <0

<=> ab +5(a+ b)+25 <0

<=> (m-5) +5.(m-1)+25 <0

<=>6m +15 <0 => m <15/6