1/ Hàm số đi qua điểm \( A(1;2)\)
\(→2=m.1-2\\↔4=m\)
2/ \(\dfrac{1}{2}x^2-mx+2=0\\↔x^2-2mx+4=0\\Δ'=(-m)^2-1.4=m^2-4=(m-2)(m+2)\)
Pt có 2 nghiệm phân biệt
\(→Δ'=(m-2)(m+2)>0\\↔\left[\begin{array}{1}\begin{cases}m-2>0\\m+2>0\end{cases}\\\begin{cases}m-2<0\\m+2<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}m>2\\m>-2\end{cases}\\\begin{cases}m<2\\m<-2\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}m>2\\m<-2\end{array}\right.\)
Theo Vi-ét:
\(\begin{cases}x_1+x_2=2m\\x_1x_2=4\end{cases}\)
\( (x_1+1)^2+(x_2+1)^2=18\\↔x_1^2+2x_1+1+x_2^2+2x_2+1=18\\↔(x_1^2+2x_1x_2+x_2^2)+2(x_1+x_2)-2x_1x_2=16\\↔(x_1+x_2)^2+2(x_1+x_2)-2x_1x_2=16\\↔(2m)^2+2.2m-2.4=16\\↔4m^2+4m-8=16\\↔4m^2+4m-24=0\\↔m^2+m-6=0\\↔m^2+3m-2m-6=0\\↔m(m+3)-2(m+3)=0\\↔(m-2)(m+3)=0\\↔\left[\begin{array}{1}m-2=0\\m+3=0\end{array}\right.\\↔\left[\begin{array}{1}m=2(KTM)\\m=-3(TM)\end{array}\right.\)
Vậy \(m=-3\)
