Bài 2 ys2
a/Thay m=-1 vào pt(1) ta đc pt ms:
x²+2x-8=0
<=> (x+4).(x-2)=0
<=>\(\left[ \begin{array}{l}x1=-4\\x2=2\end{array} \right.\)
b/ AD Viet có: $\left \{ {{x1+x2=-b/a=2m} \atop {x1.x2=c/a=4m-4}} \right.$
<=>x1²+(x1+x2).x2=12
<=>x1²+x1.x2+x2²=12
<=>(x1²+x2²)+x1.x2 =12
<=>(x1+x2)²-x1.x2=12
<=>(2m)²-(4m-4)=12
<=>4m²-4m-8=0
<=>(m+1)(m-2)=0
<=>$\left \{ {{m=2} \atop {m=-1}} \right.$
phần c mik chịu
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