已知x²-2x-8=0,求代数式(x-1)+(x+3)(x-3)+(x-3)(x-1)的值

2024-12-16 13:27:49
推荐回答(2个)
回答1:

x²-2x-8=0
(x-4)(x+2)=0
x=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

回答2:

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