关于一元二次方程的问题，配方法

-3x^2-2x+1
=-(x^2+2x/3)+1
=-[x^2+2*x*(1/3)+(1/3)^2-(1/3)^2]+1
=-(x+1/3)^2+(1/3)^2+1
=-(x+1/3)^2+10/9

x^2-(1/2)x+1
=x^2-2x*(1/4)+(1/4)^2-(1/4)^2+1
=(x-1/4)^2+15/16

(2/3)y^2+(1/3)y-2
=(2/3)(y^2+y/2)-2
=(2/3)[y^2+2*y*(1/4)+(1/4)^2-(1/4)^2]-2
=(2/3)(y+1/4)^2-(2/3)*(1/4)^2-2
=(2/3)(y+1/4)^2-49/24

ax^2+bx+c(a≠0）
=a(x^2+bx/a)+c
=a[x^2+2x*(b/2a)+(b/2a)^2-(b/2a)^2]+c
=a(x+b/2a)^2-a*(b/2a)^2+c
=a(x+b/2a)^2+(4ac-b^2)/4a

-3x²-2x+1 =-3（x²+2/3x+1/9-1/9)+1=-3(x+1/3)²+4/3
x²-(1/2)x+1 =x²-(1/2)x+1/16-1/16+1=(x-1/4)²+15/16
(2/3)y²+(1/3)y-2 =2/3(y²+1/2y)-2=2/3(y²+1/2y+1/16-1/16)-2=2/3(y+1/4)²-49/24
ax²+bx+c(a≠0）=a(x²+b/ax)+c=a(x²+b/ax+b²/4a²-b²/4a²)+c=a(x²+b/2a)²+(4ac-b^2)/4a

-3x^2-2x+1=-3(x+1/3)^2+4/3

x²-(1/2)x+1=(x-1/4)^2+15/16

(2/3)y²+(1/3)y-2