a)Thay x=1 ta có:
1+m.1-4.1-4=0
<=>m-7=0
<=>m=7
b)Với m=7 ta có:
x3+7x2-4x-4=0
<=>(x3-x2)+(8x2-8x)+(4x-4)=0
<=>(x-1)(x2+8x+4)=0
=>x2+8x+4=0
<=>x2+8x+16-12=0
<=>(x+4)2=12
<=>x+4=\(^+_-\sqrt{12}\)
<=>x=\(\sqrt{12}\)-4 hoặc x=\(-\sqrt{12}-4\)
Vậy...