已知(x y)的²=1,(x-y)的²=17

原式=(y-x)/xy=-xy/xy=-1

移项,x=y+xy:同除y得x/y=1+x;同除X得1/y=1/x+1;移项1/x-1/y=-1

1/x+1/y=3∴x+y=3xy3x-2xy+3y/x+xy+y=[3(x+y)-2xy]/[(x+y)+xy]=(9xy-2xy)/(3xy+xy)=7/4

代入方法：分母：3x-3xy+3y=3(x+y)-3xy=3•(3xy)-3xy=9xy-3xy=6xy分子：x+xy+y=x+y+xy=3xy+xy=4xy所以原式=6xy/4xy=3/

已知1/x-1/y=3,应该是1/x+1/y=3,[按原式得不出结果!]如此,则3xy＝x+y（3x-xy+3y）/（x+xy+y）＝2

1/x-1/y=(y-x)/xy=-3y=x=-3xyx-y=3xy所以原式=[5(x-y)+xy]/(x-y)-xy]=[5(3xy)+xy]/[(3xy)-xy]=16xy/2xy=8

令F（XY）=1/XY+XY,当XY=1的时候,F（XY)=2,最小.（可由函数图形象得出）.XY趋于正无穷大的时候F(XY)趋于正无穷大,XY无限趋于零的时候F(XY)趋于正无穷大.所以XY越接近1

=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10

首先化简通分1/x+1/y=5得：（x+y)/xy=5即有x+y=5xy将要求的分式化简：3x+xy+3y/x-2xy+y=3x+3y+xy/x+y-2xy（根据加法交换律）=3（x+y)+xy/(x

因为1/x+1/y=3所以x+y=3xy所以x+xy+y/2x-3xy+2y=（3xy+xy)/(6xy-3xy)=4xy/3xy=4/3

1/x+1/y=5(x+y)/x*y=5x+y=5*x*yx+2xy+y=x+y+2xy=5xy+2xy=7xy2x-3xy+2y=10xy-3xy=7xy(x+2xy+y)/(2x-3xy+2y)=

1/x-1/y=3(y-x)/xy=3y-x=3xyx-y=-3xy(5x+xy-5y)/(x-xy-y)=[5(x-y)+xy][(x-y)-xy]=(-15xy+xy)/(-3xy-xy)=-14

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15

答：x+y=-1,xy=-2-5（x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)

5xy+4x+7y+6x-3xy-4xy+3y=5xy-3xy-4xy+4x+6x+7y+3y=（5-3-4）xy+（4+6）x+（7+3）y=-2xy+10x+10y=-2xy+10（x+y）因为x

是求(3x+xy+3y)/(x-xy+y)吧?(3x+xy+3y)/(x-xy+y)上下同除以xy=(3/y+1+3/x）/(1/y-1+1/x)=[3(1/x+1/y)+1]/(1/x+1/y-1)

-5（x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4（x+y）+2xy=-4×（-1）+2×（-2）=4+（-4）=0你有问题也可以在这里向我提问

解题思路：：∵x+y=0,x+13y=1，解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程：已知x+

这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采