急!急!急!已知x²-2x-1=0,求(2x-1)²-x(x+4)+(x-2)(x+2)的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 08:02:44
![急!急!急!已知x²-2x-1=0,求(2x-1)²-x(x+4)+(x-2)(x+2)的值](/uploads/image/z/15255341-53-1.jpg?t=%E6%80%A5%21%E6%80%A5%21%E6%80%A5%21%E5%B7%B2%E7%9F%A5x%26%23178%3B-2x-1%3D0%2C%E6%B1%82%EF%BC%882x-1%EF%BC%89%26%23178%3B-x%EF%BC%88x%2B4%EF%BC%89%2B%EF%BC%88x-2%EF%BC%89%EF%BC%88x%2B2%EF%BC%89%E7%9A%84%E5%80%BC)
急!急!急!已知x²-2x-1=0,求(2x-1)²-x(x+4)+(x-2)(x+2)的值
急!急!急!已知x²-2x-1=0,求(2x-1)²-x(x+4)+(x-2)(x+2)的值
急!急!急!已知x²-2x-1=0,求(2x-1)²-x(x+4)+(x-2)(x+2)的值
原式
=4x平方-4x+1-x平方-4x+x平方-4
=4x平方-8x-3
=4(x平方-2x-1)+4-3
=4-3
=1