求证:多项式(a—2)(a^2+2a+4)—[3a(a+1)^2-2a(a-1)^2-(3a+1)(3a-1)]+a(1+a)的值与a的取值无关

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/02 07:15:09

求证:多项式(a—2)(a^2+2a+4)—[3a(a+1)^2-2a(a-1)^2-(3a+1)(3a-1)]+a(1+a)的值与a的取值无关
求证:多项式(a—2)(a^2+2a+4)—[3a(a+1)^2-2a(a-1)^2-(3a+1)(3a-1)]+a(1+a)的值与a的取值无关

求证:多项式(a—2)(a^2+2a+4)—[3a(a+1)^2-2a(a-1)^2-(3a+1)(3a-1)]+a(1+a)的值与a的取值无关
(a—2)(a^2+2a+4)—[3a(a+1)^2-2a(a-1)^2-(3a+1)(3a-1)]+a(1+a)
=a^3-8-[3a^3+6a^2+3a-(2a^3-4a^2+2a)-9a^2+1]+a+a^2
=a^3-8-[a^3+a^2+a+1]+a+a^2
=a^3-8-a^3-a^2-a-1+a+a^2
=-7
因为是恒值.所以值与a的取值无关