2x y=4,x 3z=1

(1)[4(x²y²-2xy+1)-(2xy-x²y²+4-2xy)]÷1/4xy=(4x²y²-8xy+4-2xy+x²y

f(x,y)={xy/[2-√(4+xy)]=-2-√（4+xy),xy≠0；{4,xy=0,在点(0,0),(1,0)处不连续,在（1,2）处连续.再问：能简述下原因么？再答：f(0+,0+)=-4

原式=2xy^2-4y+xy-xy-3xy^2+4y=-xy^2=-4×（1/2）^2=-1

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

令t=xy方程变为：[(t-2)^2+4t-4]/(-t)=(t^2-4t+4+4t-4)/(-t)=-t=-xy又x=10,y=-1/25-xy=-10*(-1/25)=2/5

[（xy-2)²+4xy-4]÷(xy)=(x²y²-4xy+4+4xy-4)÷(xy)=x²y²÷xy=xy=10×1/25=2/5

可以先设xy=t原式=[4(t-1)^2-(t+2)(2-t)]/(1/4*t)=4(4t^2-8t+4-2t+t^2-4+2t)/t=4(5t^2-8t)/t=(20t-32)/t=20-32/t将

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

化简结果是：-20xy+32代入得40

{（xy-2）^2+4xy-4}/(-xy)={(xy-2)^2+2(xy-2)+1+2xy-1}/(-xy)={[(xy-2)+1]^2+2xy-1}/(-xy)={(xy-1)^2+2xy-1}/

(-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*

xx+2xy-yy=-3两式相加即可

原式=(4x²y²-8xy+4-4+x²y²)÷(1/4xy)=(5x²y²-8xy)÷(1/4xy)=20xy-32=20*(-2)*(1

[4(xy-1)^2-(xy+2)(2-xy)]除以4分之1xy=[4x^2乘以y^2-8xy+4-(4-x^2y^2)]除以4分之1xy=(4x^2乘以y^2-8xy+4-4+x^2y^2)除以4分

[4(xy-1)^2-(xy+2)(2-xy)]1/4xy=[4(x²y²-2xy+1)-(4-x²y²)]1/4xy=(4x²y²-8xy

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

3x^2y-[2xy^2-4(1/2xy-3/4x^2y)+xy]+3xy^2=3x^2y-2xy^2+2xy-3x^2y-xy+3xy^2=xy^2+xy由x=3,y=-1/3得：原式=3*1/9-

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