Câu 3 :
\(x^2-2x+m-2=0\)
\(a,\) Hệ số :
\(a=1\)\(,b=-2,c=m-2\)
\(b,m=-1\Rightarrow x^2-2x-1-2=0\)
\(\Rightarrow x^2-2x-3=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(c,\) Để pt có 2 nghiệm \(x_1,x_2\) thì \(\Delta>0\Leftrightarrow\left(-2\right)^2-4\left(m-2\right)>0\Leftrightarrow4-4m+8>0\Rightarrow-4m+12>0\Rightarrow m< 3\)
Theo Vi-ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=m-2\end{matrix}\right.\)
Ta có : \(3\left(x_1^2+x_2^2\right)+x_1^2x_2^2=11\)
\(\Leftrightarrow3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+\left(x_1x_2\right)^2=11\)
\(\Leftrightarrow3\left(2^2-2\left(m-2\right)\right)+\left(m-2\right)^2=11\)
\(\Leftrightarrow3\left(4-2m+4\right)+m^2-4m+4-11=0\)
\(\Leftrightarrow12-6m+12+m^2-4m+4-11=0\)
\(\Leftrightarrow m^2-10m+17=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=5+2\sqrt{2}\left(ktm\right)\\m=5-2\sqrt{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(m=5-2\sqrt{2}\) thì thỏa mãn đề bài.
