(1)
(x²-x+1)²
= (x²-x)²+2(x²-x)+1²
=x⁴-2x³+x²+2x²-2x+1
=x⁴-2x³+3x²-2x+1
(2)
(3m+n-p)²
=(3m)²+6m(n-p)+(n-p)²
=9m²+6mn-6mp+n²-2np+p²
(3)
(x+y)(x-y)(x²+y²)
= (x²-y²)(x²+y²)
=x⁴-y⁴
(4)
(x-y+z)-(x+y-z)²
前面个括号是否漏掉了一个平方符号
(5)
(1-a)(a+1)(a²+1)(a的四次方+1)
=-(a²-1)(a²+1)(a⁴+1)
=-(a⁴-1)(a⁴+1)
=-(a⁸ -1)
=1-a⁸
(6)
[xy+(z+m)][xy-(z+m)]
=(xy)²-(z+m)²
=x²y²-z²-2mz-m²
(x²-x+1)²= (3m+n-p)²= (x+y)(x-y)(x²+y²)= (x-y+z)-(x+y-z)²=
(x²-x+1)²= (3m+n-p)²= (x+y)(x-y)(x²+y²)= (x-y+z)-(x+y-z)²=
(1-a)(a+1)(a²+1)(a的四次方+1)=
[xy+(z+m)][xy-(z+m)]=
请用初二上的知识回答,谢谢【鞠躬】
(1-a)(a+1)(a²+1)(a的四次方+1)=
[xy+(z+m)][xy-(z+m)]=
请用初二上的知识回答,谢谢【鞠躬】
其他人气:774 ℃时间:2020-09-09 18:17:17
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