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  • 化簡求值[1-(2x/x+y)]÷ (x²-2xy+y²/3x+3y)+(x²+xy/x²-y²)其中x=-7,y=6

    化簡求值[1-(2x/x+y)]÷ (x²-2xy+y²/3x+3y)+(x²+xy/x²-y²)其中x=-7,y=6
    數(shù)學(xué)人氣:152 ℃時間:2019-08-20 12:16:23
    優(yōu)質(zhì)解答

    1、化簡
    [1-(2x/x+y)]÷ (x²-2xy+y²/3x+3y)+(x²+xy/x²-y²)
    =[(x+y)/(x+y)-(2x)/(x+y)]÷[(x-y)²/3(x+y)]+[x(x+y)/(x+y)(x-y)]
    =[(x+y-2x)/(x+y)]÷[(x-y)²/3(x+y)]+x/(x-y)
    =[-(x-y)/(x+y)][3(x+y)/(x-y)(x-y)]+x/(x-y)
    =-3/(x-y)+x/(x-y)
    =(x-3)/(x-y)
    2、求值

    x=-7,y=6

    原式=(x-3)/(x-y)
    =(-7-3)/(-7-6)
    =(-10)/(-13)
    =10/13
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