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  • 計(jì)算:(1)(4x-1)(x+4) (2)(-2x-1/2)(x-1) (3)(x-2)(x+3)-2x²

    計(jì)算:(1)(4x-1)(x+4) (2)(-2x-1/2)(x-1) (3)(x-2)(x+3)-2x²
    (4)(a+1)(a-1/2)
    (5)(3x-2y)(3x+2y)
    (6)(3x-2)²
    (7)(2x-5y)(3x-5y+6)
    數(shù)學(xué)人氣:786 ℃時(shí)間:2020-02-02 21:28:08
    優(yōu)質(zhì)解答
    (1)(4x-1)(x+4) =4x²+16x-x-4=4x²+15x-4
    (2)(-2x-1/2)(x-1) =-2x²+2x-1/2x+1/2=-2x²+3/2x+1/2
    (3)(x-2)(x+3)-2x²=x²+x-6-2x²=-x²+x-6
    (4)(a+1)(a-1/2) =a²-1/2a+a-1/2=a²+1/2a-1/2
    (5)(3x-2y)(3x+2y)=(3x)²-(2y)²=9x²-4y²
    (6)(3x-2)²=(3x)²-2×3x×2+2²=9x²-12x+4
    (7)(2x-5y)(3x-5y+6)=6x²-10xy+12x-15xy+25y²-30y=6x²-25xy+25y²+12x-30y
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