问题
解答题
已知x2-4x-1=0,求代数式(2x-3)2-(x+y)(x-y)-y2的值.
答案
原式=4x2-12x+9-x2+y2-y2
=3x2-12x+9
=3(x2-4x+3),
∵x2-4x-1=0,即x2-4x=1,
∴原式=12.
已知x2-4x-1=0,求代数式(2x-3)2-(x+y)(x-y)-y2的值.
原式=4x2-12x+9-x2+y2-y2
=3x2-12x+9
=3(x2-4x+3),
∵x2-4x-1=0,即x2-4x=1,
∴原式=12.