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  • (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)]=
    請用初二上的知識回答,謝謝【鞠躬】
    其他人氣:652 ℃時間:2020-09-09 18:17:17
    優(yōu)質(zhì)解答
    (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?zhèn)€括號是否漏掉了一個平方符號
    (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
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