已知x-3y=0求4x

x^2-4x+y^2-10y+29=0(x^2-4x+4)+(y^2-10y+25)=0(x-2)^2+(y-5)^2=0x=2,y=5x^2y^2+2x^3y^2+x^4y^2=x^2y^2(1+2

(2x+y)/(x^2-2xy+y^2)X(x-y)=(2x+y)/（x-y）²*（x-y)=(2x+y)/（x-y）x-3y=0,x=3y(2x+y)/（x-y）=（6y+y）/（3y-y

x^2-4x+y^2-10y+29=(x^2-4x+4)+(y^2-10y+25)=(x-2)^2+(y-5)^2=0所以x=2,y=5因此X^2Y^2+2X^3Y^2+X^4Y^2=4*25+2*8

x平方+y平方-4x+6y+13=0(x-2)^2+(y+3)^2=0x=2,y=-3;带入要求的表达式即可.再问：怎么算出x=2y=-3的？？？再答：例如：a^2+b^2=0就只有a=0,b=0才成

4x-2y=0→y=2x,把所有的y都用2x替代.结果是-1/7

x*x+y*y-4x-6y+13=0可以化为：(x-2)^2+(y-3)^2=0只有在x=2、y=3时,上式才能成立.代入3x+4y=3*2+4*3=18

x^2-4x+29-10y+y^2=(x-2)^2+(y-5)^2=0x=2y=5x^2y^2+2x^3y^2+x^4y^2=4*25+2*8*25+16*25=900

x^2+3xy-4y^2/y^2=0x/y=-4or1x-y/x+y=5/3or0

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：{

X²+3XY-4Y²=0（x+4y）（x-y）=0x=y或x=-4y若x=y,则X-Y/X+Y=0若x=-4y,则X-Y/X+Y=-5y/(-3y)=5/3

（x+2）^2+|x+y+5|=0,x+2=0,x+y+5=0x=-2,y=-33(x^2)y-2(x^2)y+2xy-(x^2)y+4x^2-xy=4x^2+xy=4*(-2)^2+(-2)*(-3

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

|3-y|+|x+y|=0，且|3-y|≥0，|x+y|≥0，所以3-y=0，x+y=0，所以y=3，x=-3．所以x+yxy=-3+3-3×3=0-9=0．答：x+yxy的值为0．

(x³+y³)+(x²y+xy²)+(x+y)=0(x+y)(x²-xy+y²)+xy(x+y)+(x+y)=0(x+y)(x²-

∵绝对值≥0,平方≥0即|x+1|≥0,(3-4y-x)²≥0∴|x+1|+(3-4y-x)²=0,只能是|x+1|=0且(3-4y-x)²=0∴由|x+1|=0可知x=

2x＋y＋3=xy,①若x=1,则y不存在,此时5x＋4y无最值；②若x≠1,则y=(2x＋3)/(x－1),M=5x＋4y=5x＋(8x＋12)/(x－1)=(5x²＋3x＋12)/(x－

X*2+Y*2=(X+Y)*2__-2XY;因为X+Y=3;XY=2;所以X*2+Y*2=9-4=5.

x+2=0,x+y+5=0x=-2,y=-33x^2y+{-2X^2y-[-2xy+(x^2y-4x^2)]-xy}=3x^2y-2X^2y+2xy-x^2y+4x^2-xy=4x^2+xy=4*4+

丨x+3丨+丨y-x+4丨=0丨x+3丨=丨y-x+4丨=0x+3=0,y-x+4=0x=-3,y=-7(x-y)(x+y)=4*(-10)=-40

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.