解2(X Y)dx YDY=0-搜答案-拍题作业网

y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5

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×(

参考一下

解1式：(x+4y)(x-y)=0x=-4y时,解2式得：x=-10/3,y=5/6;x=10/3,y=-5/6x=y时,解2式得：x=5,y=5;x=-5,y=-5

x²-2xy+y²=12x²-5xy-3y²=0即(x-y)²=1(2x+y)(x-3y)=02x+y=0或x-3y=0如果2x+y=0,则y=-2x

因为x+2y=0,所以x=-2y原式=(x^2+2xy)/(xy+y^2)=(4y^2-4Y^2)/(-3y^2+y^2)=0/(-2y^2)又因为xy不等于零,所以x、y君不等于零,所以-2y^2亦

伯努利方程xy'+y=2y^3->x/y^3*y'+1/y^2=2令1/y^2=t-x/2*dt/dx+t=2解这个一阶方程得(2x^(-2)+c)*x^2

两式相减x²-10x-y²+10y=0x²-y²-10(x-y)=0(x+y)(x-y)-10(x-y)=0(x+y-10)(x-y)=0x+y=10x-y=0

x+2y=0,xy不等于0∴x=-2yx²+2xy/xy+y²=(4y²-4y²)/(-2y²+y²)=0

做边量替换,u=y/x,即y=uxy’=u+xu'原方程左右同除x^2y变为（1-u+u^2)+(1/u+1+u)(u+xu')=0积分再换回变量就是答案了不知道你会不会积分,再问：还是写下过程吧，没

y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5

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

xy+2x+y=0y(x+1)+2x+2=2y(x+1)+2(x+1)=2(2+y)(x+1)=2因为x,y都是整数因此2+y=1,x+1=2,或2+y=-1,x+1=-2或2+y=2,x+1=1或2

根据题意，2x2-3xy+y2=0，且xy≠0，故有(yx)2−3yx+2＝0，即(yx−1)(yx−2)＝0，即得yx=1或2，故xy=1或12，所以xy+yx=2或212．故选A．

4种x=-3,y=4x=-1,y=4x=1,y=-4x=3,y=-4

x²+2xy+y²=4x²-xy-y²=0方程1化为(x+y)²=4,得x+y=2或x+y=-2将x=2-y代入方程2得：4-4y+y²-2

x+y=5xy（2x-3xy+2y)/(x+2xy+y）=[2（x+y）-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问：若x＋1/x＝3,求(x

x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式＝9y²-3y²+y²/6y²=7/6x=4y时原式