已知x²-4x-1=0,求代数式(2x-3)²-(x+y)(x-y)-y²的值

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已知x²-4x-1=0,求代数式(2x-3)²-(x+y)(x-y)-y²的值

因为x^2-4x-1=0
所以x^2-4x=1
原式=4x^2-12x+9-x^2+y^2-y^2
=3x^2-12x+9
=3(x^2-4x)+9
=3*1+9
=3+9
=12

(2X-3)²-(X+ Y)(X-Y)-Y²=4X²-12X+9-(X²-Y²)-Y²=3X²-12X+9
=3(X²-4X)+9=3乘1+9=12