a, Có : \(\left\{{}\begin{matrix}a=1\\b^,=-\left(m-5\right)\\c=1-4m\end{matrix}\right.\)
=> \(\Delta^,=b^{,^2}-ac=\left(-\left(m-5\right)\right)^2-\left(1-4m\right)\)
=> \(\Delta^,=m^2-10m+25-1+4m\)
=> \(\Delta^,=m^2-6m+24=m^2-2.m.3+9+15=\left(m-3\right)^2+15\)
Ta thấy \(\Delta^,\ge15>0\)
Nên phương trình có 2 nghiệm phân biệt với mọi m .
b, Theo vi ét : \(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=2\left(m-5\right)=2m-10\\x_1x_2=\frac{c}{a}=1-4m\end{matrix}\right.\)
- Để \(2x_1^2+x_1^2x_2+x_2^2x_1+2x^2_2=6\)
<=> \(2\left(x_1^2+x_2^2\right)+x_1x_2\left(x_1+x_2\right)=6\)
<=> \(2\left(\left(x_1+x_2\right)^2-2x_1x_2\right)+x_1x_2\left(x_1+x_2\right)=6\)
<=> \(2\left(\left(2m-10\right)^2-2\left(1-4m\right)\right)+\left(1-4m\right)\left(2m-10\right)=6\)
<=> \(2\left(4m^2-40m+100-2+8m\right)+\left(2m-8m^2-10+40m\right)-6=0\)
<=> \(8m^2-80m+200-4+16m+2m-8m^2-10+40m-6=0\)
<=> \(180-22m=0\)
<=> \(m=\frac{90}{11}\)
( sửa đề roài nhoa )
