方程x^2y-xy-4y=x^3a的通解是

y(x^2-xy+y^2)dx=-x（x^2-xy+y^2）dy,当y≠0时,x^2-xy+y^2=(x-0.5y)^2+3/4y^2＞0,两边约去此式,得ydx=-xdy,-dx/x=dy/y,易得

y(x^2-xy+y^2)dx＋x（x^2-xy+y^2）dy=0(x^2-xy+y^2)(dxy+dxy)=0(x^2-xy+y^2)*2dxy=02dxy=0(1)或者x^2-xy+y^2=0(2

x^2+(4-3y)x+3y^2-18y+52=0x=0.5(3y-4+sqrt((4-3y)^2-4(3y^2-18y+52)))=0.5(3y-4+sqrt(-3(y-8)^2))或x=0.5(3

这是哪里的题,你确认没有写错?再问：没有啊，这是高一衔接题目再答：借鉴迷路的猫猫原式以y为主元，因式分解，可得：2y²+(4+x)y-x²-5x-6=02y²+(4+x)

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

设y/x=k,即有y=kx代入方程中有：x^2+k^2x^2-4x+1=0(1+k^2)x^2-4x+1=0判别式=16-4(1+k^2)>=01+k^2

原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1

(x-2y)^2+(y-3)^2=0y=3x=6

x^4-x^2+2xy+y^2+1=0x^4-2x^2+1+x^2+2xy+y^2=0(x^2-1)^2+(x+y)^2=0因为(x^2-1)^2>=0,(x+y)^2>=0当且仅当(x^2-1)^2

x²+2xy+6x+2y²+4y+10=0x²+2xy+y²+6x+6y+9+y²－2y+1=0(x+y)²+6（x+y）+9+y²

(x-y)^2=9,(x+2y)^2+(x+2y)=2,2x-1/y=3[2xy-1=3y],x=2y[4y^2-3y-1=0,y=1或者y=-1/4],x=2或者x=-1/2

x^2y^2-8xy+4x^2+y^2+4=0(x^2y^2-4xy+4)+(4x^2-4xy+y^2)=0(xy-2)^2+(2x-y)^2=0平方相加等于0则都等于0所以xy=2,2x=y所以x*

3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出：{

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

（1）、x^2+2xy+y^2-4x-4y-5=(x+y-5)(x+y-1)=0,——》x+y=5,或x+y=-1,x^2-2xy+y^2-2x+2y-3=(x-y-3)(x-y+1)=0,——》x-

令y=kxx*x+kx*x=k*x+k*k*x*x(1-k*k+k)x^2-kx=0x((1-k*k+k)x-k)=0由上式得X=0或(1-k*k+k)x-k=0解得：k=(x-1+（或-）√（（1-

因为2xy+x+y=83所以4xy+2x+2y=83*2所以(2x+1)(2y+1)=165=3*5*11=1*165=3*55=5*33=11*15所以2x+1=1,2y+1=165或2x+1=3,

[-(x-y)^2/xy]^-4*(y^2-xy/x)^3*(x^4/y^10)÷(xy-y^2/x)^-5=-y(y-x)^29^-x*27^y=3^-2x*3^3y=3^(-2x+3y)=3^(-

4x^2-4xy+2y^2+2y+1=0(4x^2-4xy+y^2)+(y^2+2y+1)=0(2x-y)^2+(y+1)^2=02x-y=0,y+1=0x=-1/2,y=-1