题目

不然明天考试就被拒了!帮我把列式过程讲清楚了啊.
[a-b+c][a+b-c];这题老师列式是这样,理解不来；a^2-[b-c]^2
[a+b-c][a+b+c]
[x-y]^2[y-x][x-y]^3[y-x]^2
当k=-2时,多项式x-1与2-kx的乘积不含一次项

提问时间：2021-02-28

答案

[a-b+c][a+b-c]=[a-(b-c)][a+(b-c)]=a^2-(b-c)^2
[a+b-c] [a+b+c]=[(a+b)-c)][(a+b)+c]=(a+b)^2-c^2
[x-y]^2[y-x][x-y]^3[y-x]^2=(x-y)^2[-(x-y)](x-y)^3[-(x-y)]^2=-(x-y)^2(x-y)(x-y)^3(x-y)^2
=-(x-y)^(2+1+3+2)=-(x-y)^8
k=-2代入(x-1)(2-kx)得:
(x-1)(2-kx)=(x-1)[2-(-2)x]=(x-1)(4x)=4x^2-4x,因不含一次项,所以4x=0,x=0