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  • 計(jì)算: (1)x2-(x3y-2x2y2)﹢xy (2)(1-1/2)×(1+1/2)×(1+1/4)×(1+1/16) (3)(2x-3y)2-(y+3x)(3x-y) (4)(2a3b2-4a4b3+6a5b4)÷(-2a3b2)

    計(jì)算:
    (1)x2-(x3y-2x2y2)﹢xy
    (2)(1-
    1
    2
    )×(1+
    1
    2
    )×(1+
    1
    4
    )×(1+
    1
    16

    (3)(2x-3y)2-(y+3x)(3x-y)
    (4)(2a3b2-4a4b3+6a5b4)÷(-2a3b2
    數(shù)學(xué)人氣:753 ℃時(shí)間:2020-04-04 07:16:48
    優(yōu)質(zhì)解答
    (1)原式=x2-x3y+2x2y2+xy;
    (2)原式=(1-
    1
    4
    )×(1+
    1
    4
    )×(1+
    1
    16
    )=(1-
    1
    16
    )(1+
    1
    16
    )=1-
    1
    256
    =
    255
    256
    ;
    (3)原式=4x2-12xy+9y2-9x2+y2=-5x2-12xy+10y2;
    (4)原式=-1+2ab-3a2b2
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