1、已知2x-y=8,求[(x^2+y^2)-(x-y)^2+2y(x-y)]除以(4y)的值.

1个回答

  • 1,化简[(x^2+y^2)-(x-y)^2+2y(x-y)]除以(4y),即:(2xy+2xy-2y^2)/(4y)=(2x-y)2=4

    2,原式=1.2345^2+2×1.2345×0.7655+0.7655^2=(1.2345+0.7655)^2=4

    3,原式可等价化为:(m+2)^2+(n-3)^2=0,所以m=-2,n=3,所以m^2-n^2=-5

    4,x^2-6x+b配方可得:(x-3)^2+b-9与(x-a)^2-1等价,所以a=3,b-9=-1,b=8,所以b-a=5

    5,原式等价于a^2b^2-2ab+1+a^2+b^2-2ab=0,即(ab-1)^2+(a-b)^2=0,所以ab=1,a-b=0

    6,a^2-6a+9与|b-1|互为相反数,所以a^2-6a+9=-|b-1|,即(a-3)^2+|b-1|=0,所以a=3,b=1,

    b分之a-a分之b)除以(a+b)的值为2/3

    7,x-y=a^2+4a+4+b^2-8b+16=(a+2)^2+(b-4)^2>=0,所以x>=y