已知x²-2x-8=0,求代数式(x-1)+(x+3)(x-3)+(x-3)(x-1)的值
问题描述：
已知x²-2x-8=0,求代数式(x-1)+(x+3)(x-3)+(x-3)(x-1)的值
答
知x²-2x-8=0,
(x-4)(x+2)=0
x=4 或x=-2
代数式(x-1)+(x+3)(x-3)+(x-3)(x-1)的值
=(x-1)+(x+3+x-1)(x-3)
=(x-1)+(2x+2)(x-3)
=x-1+2x^2-6x+2x-6
=2x^2-3x-7
(1)x=4
上式=2*4^2-3*4-7=13
(2)x=-2
上式=2*(-2)^2-3*(-2)-7=7
答
x²-2x-8=0(x-4)(x+2)=0x=4,x=-2(x-1)+(x+3)(x-3)+(x-3)(x-1)=x-1+x²-9+x²-4x+3=2x²-3x-7当x=4时,原式=2×4²-3×4-7=13当x=-2时,原式=2×(-2)²-3×(-2)-7=7