Question

已知x^2-4x+4=0.2y^2+12y+18=0,求代数式9x^2-12xy+4y^2的值

Answer 1

x^2-4x+4=0

(x-2)^2=0

x=2

2y^2+12y+18=0

2(y^2+6y+9)=0

2(y+3)^2=0

y=-3

9x^2-12xy+4y^2

=(3x-2y)^2

=[3×2-2×（-3）]^2

=12^2=144

m^6-1

=(m^3+1)(m^3-1)

=(m-1)(m^2+m+1)(m+1)(m^2-m+1)