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  • (x的2方+1)(x+1)(x-1)=要過程

    (x的2方+1)(x+1)(x-1)=要過程
    (2a+1/2)(2a-1/2)(4a的2方+1/4)=
    (x+3)(x-3)-2(2-x)(2+x)=
    (2x+3y)(3x-2y)-(x-2y)(x+2y)=
    (a的2方-4)(a的2方-5)+(3a+5)(3a-5)=
    [(2a-1)(2a+1)+(a+2)(a-2)-4a的2方](a的2方+5)=
    數(shù)學人氣:171 ℃時間:2020-06-09 19:00:17
    優(yōu)質(zhì)解答
    (x^2+1)(x+1)(x-1)
    =(x^2+1)(x^2-1)
    =x^4-1
    (2a+1/2)(2a-1/2)(4a^2+1/4)
    =(4a^2-1/4)(4a^2+1/4)
    =16a^4-1/16
    (x+3)(x-3)-2(2-x)(2+x)
    =x^2-9-2(4-x^2)
    =x^2-9-8+2x^2
    =3x^2-17
    (2x+3y)(3x-2y)-(x-2y)(x+2y)
    =6x^2+5xy-6y^2-x^2+4y^2
    =5x^2+5xy-2y^2
    (a^2-4)(a^2-5)+(3a+5)(3a-5)
    =a^4-9a^2+20+9a^2-25
    =a^4-5
    [(2a-1)(2a+1)+(a+2)(a-2)-4a^2](a^2+5)
    =(4a^2-1+a^2-4-4a^2)(a^2+5)
    =(a^2-5)(a^2+5)
    =a^4-25
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