(X y)^2=1,x^2 xy=3,求2x^2-XY-3Y^2的值

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

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

[（xy+2）（xy-2）-2x2y2+4]÷（xy）=（x2y2-4-2x2y2+4）÷xy=-x2y2÷xy=-xy,当x=10,y=1/25时,原式=-2/5．

[(xy+2)(xy-2)-2x^2y^2+4]/(xy)=[(x^2y^2-4)-2x^2y^2+4]/(xy)=-x^2y^2/xy=-xy=-10*(-1/25)=2/5

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

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

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2

(8xy-3x平方)-5xy-2(3xy-2x平方)=8xy-3x²-5xy-6xy+4x²=x²-3xy=4+6=10

原式=3x-3y-6xy=3-12=-9

（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

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

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3（x+y）+2xy=3*（-2）+2*3=0

(3x^2+2xy)/xy-(2x^2-xy)/xy=(3x^2+2xy-2x^2+xy)/xy=(x^2+3xy)/xy=x(x+3y)/xy=(x+3y)/y

（x+2y-3xy)-(-2x-y+xy)=x+2y-3xy+2x+y-xy=(1+2)x+(2+1)y-(3+1)xy=3x+3y-4xy=3(x+y)-4xy=3*1/2-4*(-1/2)=3/2

[3xy(1-x)-6xy(x-1/2)]*2x*(-xy)平方=(3xy-3x²y-6x²y+3xy)(2x³y²)=12x^4y³-18x^5y&

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+

(-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