1)x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0
<=>x^3(x-1)-2x^2(x-1)+2x(x-1)-(x-1)=0
<=>(x-1)(x^3-2x^2+2x-1)=0
<=>(x-1)(x^3-x^2-x^2+x+x-1)=0
<=>(x-1)<x^2(x-1)-x(x-1)+(x-1)>=0
<=>(x-1)(x-1)(x^2-x+1)=0
<=>(x-1)^2(x^2-x+1=0
<=>(x-1)^2=0vĩ^2-x+1>0
<=>x-1=0
<=>x=1
2)nếu x=0thì 6=0(vô lí)
nẽu#0thì x^2#0
chia cả hai vế cho x^2ta được 6x^2+5x-38+5:x+6:x^2=0
<=>6(x^2+1:x^2)+5(x+1:x)-38=0(1)
đặt x+1:x=t =>t^2-2=x^2+1:x^2
(1)<=>6(t^2-2)+5t-38=0
<=>6t^2-12+5t-38=0
<=>6t^2+5t-50=0
<=>x=2,5hoặc x=-10:3
3)lam tương tự câu 2