[x y]^2=20,[x-y]^2=40求x^2 y^2,xy-搜答案-搜搜考试网

由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1

(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=40+20x+y-2xy=60(x-xy)+(xy-y)=40+(-20)x-xy+xy-y=40-20x-y=20

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

[2x(x²y-xy²)+xy(xy-x²)]/x²y=[2x^3y-2x^2y^2+x^2y^2-x^3y]/x²y=x-y=4

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

对.前提是x不等于y

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

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60希望满意!

(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60

（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

7或者-8再问：求过程^_^再答：两个等式两边相加

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

（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

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

xy+x=20 xy+y=18

由题意得：X＝Y+2.那么Y（Y+2）+Y+2＝20（Y+2）×（Y+1)=20所以y＝3那么x＝5可待入xy+y=18就不对了.（Y+2）×（Y+1)=20,Y应该是-6,X是-4,答案就对了.X＝

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

3x-4xy+3y再问：然后的过程再答：再带入值等于3.5再答：马上再答：给你照片再答：再答：亲采纳哟

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

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60