已知x-y=-1,xy=3.求x3次方y-2x²y² xy3次方的值

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

x³y+2x²y²+xy³=xy(x²+2xy+y²)=xy(x+y)²=1/2(2)²=2

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

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

绝对值大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所以x+y+1=0xy+3=0xy=-3x+y=-1两边平方x^2+2xy+y^2=(-1)^2x^2+y^2=1-

原式=2xy-6x-6y+3xy=5xy-6(x+y)=5×3-6×(-2)=27

(-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)

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)

由1/x-1/y=2001得：y-x/xy=2001所以：xy/y-x=1/2001xy/x-y=-1/2001x-xy-y/(x-y)=x-y-xy/(x-y)=1-xy/(x-y)原式=1+1/2

3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9

-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

x^3y-2x^2y^2+xy^3=xy(3x^2-4xy+y^2)=xy(3x-y)(x-y)x-y=-1,xy=3y(y-1)=3

x²+y²-xy+2x-y+1=[3（x+1）²+(x-2y+1）²]/4=0,由于（x+1）²>=0且(x-2y+1）²>=0,则有x+1

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-3xy-xy)+(2x+2x-x)+(3y-2y-4y)=-6xy+3x-3y=-6+3*3=3

(-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