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](/uploads/image/z/958924-28-4.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B-4x-1%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%EF%BC%882x-3%EF%BC%89%26%23178%3B-%EF%BC%88x%2By%EF%BC%89%EF%BC%88x-y%EF%BC%89-y%26%23178%3B%E7%9A%84%E5%80%BC)
已知x²-4x-1=0,求代数式(2x-3)²-(x+y)(x-y)-y²的值
已知x²-4x-1=0,求代数式(2x-3)²-(x+y)(x-y)-y²的值
已知x²-4x-1=0,求代数式(2x-3)²-(x+y)(x-y)-y²的值
解析
(2x-3)²-(x+y)(x-y)-y²
=4x²-12x+9-(x²-y²)-y²
=4x²-12x+9-x²+y²-y²
=3x²-12x+9
=3(x²-4x)+9
=3+9
=12
3的说法是广告