1、已知3x²+xy-2y²=0(x≠0,y≠0)求x/y-y/x-(x²+y²

4个回答

  • (1)由3x²+xy-2y²=0得:

    (x+y)(3x-2y)=0

    x=-y或 x=2y/3

    x/y-y/x-(x²+y²)/xy

    =(x²-y²)/xy-(x²+y²)/xy

    =-2y²/xy=-2y/x

    把 x=-y或 x=2y/3代入上式有

    -2y/x =2或-3

    所以x/y-y/x-(x²+y²)/xy的值的2或-3

    (2)(a+b-c)/c=(a-b+c)/b=(-a+b+c)/a

    =(a+b)/c=(a+c)/b=(b+c)/a

    =〔(a+b)+(a+c)+(b+c)〕/(c+b+a)(等式的分子分母分别相加,等式的值不变)

    =2

    所以(a+b)(b+c)(c+a)/abc

    =〔(a+b)/c〕〔(a+c)/b〕〔(b+c)/a〕

    =2×2×2

    =8