a)1+x\(\ge\)mx+m
<=>x-mx\(\ge\)m-1
<=>x(1-m)\(\ge\)m-1(1)
*)Nếu m=1 thì (1)<=>0x=0(thỏa mãn với mọi x)
*)Nếu m < 1 thì 1-m>0
(1)<=>\(x\ge\dfrac{m-1}{1-m}\)
<=>x\(\ge\)-1
*)Nếu m>1 thì 1-m<0
(1)<=>x\(\le\dfrac{m-1}{1-m}\)
<=>x\(\le-1\)
Vậy...
b)2x4-x3-2x2-x+2=0
<=>(2x4-2x3)+(x3-x2)-(x2-x)+(2x+2)=0
<=>(x-1)(2x3+x2-x+2)=0
bó tay :)