有冇人可以幫我揾個x,用m來表達:(m+1)x²-(m²+2m+2)x+m+1=0, 而m≠-1, 請列詳細步驟，thx!?

Sol
D=(m^2+2m+2)^2－4(m+1)^2
=(m^2+2m+2－2m-2)(m^2+2m+2+2m+2)
=m^2(m+2)^2
So
x=[(m^2+2m+2)+/-m(m+2)]/[2(m+1)]
(m^2+2m+2)+m^2+2m=2m^2+4m+2=2(m+1)^2
2(m+1)^2/[2(m+1)]=m+1
(m^2+2m+2)－m^2－2m=2
So
x=m+1 or x=1/(m+1)

(m+1)x²-(m²+2m+2)x+m+1=0
x={(m²+2m+2)+[(m²+2m+2)^2 -4(m+1)(m+1)]^0.5}/[2(m+1)] or x={(m²+2m+2)-[(m²+2m+2)^2 -4(m+1)(m+1)]^0.5}/[2(m+1)]
x={(m²+2m+2)+[(m²+2m+2)-2(m+1)]^0.5[(m²+2m+2)+2(m+1)]^0.5}/[2(m+1)] or x={(m²+2m+2)+[(m²+2m+2)+2(m+1)]^0.5[(m²+2m+2)+2(m+1)]^0.5}/[2(m+1)]
x={(m²+2m+2)+[(m²)(m²+4m+4)]^0.5}/[2(m+1)] or x={(m²+2m+2)-[(m²)(m²+4m+4)]^0.5}/[2(m+1)]
x=[(m²+2m+2)+m(m+2)]/[2(m+1)] or x=[(m²+2m+2)-m(m+2)]/[2(m+1)]
x=(m²+2m+2+m²+2m)/[2(m+1)] or x=(m²+2m+2-m²-2m)/[2(m+1)]
x=(2m²+4m+2)/[2(m+1)] or x=2/[2(m+1)]
x=m+1 or x=1/(m+1)