已知x²-2x-2=0,求代数式(x-1)²+(x+3)(x-3)+(x-3)(x-1)的值

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已知x²-2x-2=0,求代数式(x-1)²+(x+3)(x-3)+(x-3)(x-1)的值

(x-1)²+(x+3)(x-3)+(x-3)(x-1)
=x²-2x+1+x²-9+x²-4x+3
=3x²-6x-5
=3x²-6x-6+1
=3(x²-2x-2)+1 x²-2x-2=0
=1

x²-2x-2=0
=>(x-1)²=3
(x-1)²+(x+3)(x-3)+(x-3)(x-1)
=3+(x-3)(x+3+x-1)
=3+2(x-3)(x+1)
=3+2(x²-2x-3)
=3+2(x²-2x-2-1)
=3-2
=1