已知x的平方-x=2,求(x-2)(2x1)-(x-1)的平方-1
x²-x=2
x²-x-2=0
x=[1±√(1+8)]/2
x=(1±3)/2
x₁=2
x₂=-1
当 x=2 时
(x-2)(2x 1)-(x-1)²-1
=-(2-1)²-1
=-2
当 x=-1 时
(x-2)(2x+1)-(x-1)²-1
=(-1-2)(-2+1)-(-1-1)²-1
=-3×(-1)-4-1
=3-4-1
=-2
(x-2)(2x-1)-(x-1)²-1
=(-1-2)(-2-1)-(-1-1)²-1
=9-4-1
=4
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