已知x^2-2X=2,求代数式(x-1)^2+(X+3)(X-3)+(X-3)(x-1)的值

2024-12-05 07:17:50
推荐回答(2个)
回答1:

X^2-2X=2
X^2-2X+1=3得孝正(x-1)^2=3
(x-1)^2+(X+3)(X-3)+(X-3)(巧好悔袜源x-1)
=3+X^2-9+X^2-3X-X+3
=2X^2-9-4X+6
=2X^2-4X+2-5
=2(X^2-2X+1)-5
=2(X-1)^2-5
=1

回答2:

(x-1)^2+(x+3)(x-3)+(x-3)(x-1)
=x^2-2x+1+x^2-9+x^2-4x+3
=3x^2-6x-5
=3(x^2-2x)-5
=3*2-5
=1
注:x^2+2x=2应茄基没颤纳锋核为x^2-2x=2
(a-b)^5(b-a)^3
=-(a-b)^5(a-b)^3
=-(a-b)^8