已知x加2y=0,求分式x平方减xy-搜答案-搜搜考试网

把25拆成16+9(x²-8x+16)+(y²+6y+9)=0(x-4)²+(y+3)²=0平方大于等于0,相加等于0若有一个大于0,则另一个小于0,不成立.所

原式＝(x^2-2x+1)+(y^2-4y+4)=(x-1)^2+(y-2)^2=0,所以x=1,y=2,代入问题得结果为3/2

x-3y=0且xy≠0,所以x=3y(x≠0,y≠0)带入分式,(9y-3y+y)/(9y+3y+y)=7/13

x^2-2x+y^2+6y+10=0x^2-2x+1+y^2+6y+9=0(x-1)^2+(y+3)^2=0x-1=0,y+3=0x=1,y=-3(x-y)/(x+y)=4/(-2)=-2

x²+2x+1+y²-6y+9=0(x+1)²+(y-3)²=0∴x+1=0y-3=0∴x=-1y=3

x^2+y^2-4x-6y+13=0(x^2-4x+4)+(x^2-6y+9)=0(x-2)^2+(y-3)^2=0所以(x-2)^2=0,(y-3)^2=0x-2=0,y-3=0x=2,y=3xy/

答：x²+y²-2x+8y+17=0(x²-2x+1)+(y²+8y+16)=0(x-1)²+(y+4)²=0所以：x-1=0y+4=0解得

因为x²+4y²-4x+4y+5=x²-4x+4+4y²+4y+1=(x-2)²+(2y+1)²=0所以得x=2,y=-1/2原式=(x&#

x²-2x+y²+6y+10=0(x²-2x+1)+(y²+6y+9)=0(x-1)²+(y+3)²=0∴x-1=0,y+3=0∴x=1,y

已知x²-4xy+4y²=0即得(x-2y)²=0即得x=2y所以,(x+y)/(x-y)=(3y)/y=3

∵3X+4Y=0(X不等于0）∵y=-3x/4∴原式=（-3x²/2+9x²/16)/(x²+3x²/4)=-15/16*4/7=-105/4再问：这分式怎么X

x=-4y=-3=-516

x的平方+2xy+y的平方分之x的平方-y的平方=(x+y)(x-y)/(x+y)²=(x-y)/(x+y)∵x+y=2,x-y=1∴原式=1/2

（1）x^2+y^2-4x+10y+29=0x^2-4x+4+y^2+10y+25=0(x-2)^2+(y+5)^2=0x=2y=-5剩下的自己解下面两个题目是不是写错了,没看懂

(2x^2+3xy-2y^2)/(x^2+3xy+2y^2)={(2x-y)(x+2y)}/{(x+y)(x+2y)}=(2x-y)/(x+y)=02x-y=0

由题意可得x²+y²+2x+6y+10=0（x+1）²+（y+3）²=0所以x=-1,y=-3

因为x:y=2:5所以x/y=2//5xy/5x²+y²,分子、分母同时乘1/y²=(x/y)/(5x²/y²+1)=(2/5)/(5×4/25+1)

y/x－x/y=3,通分得到（y方-x方)/xy=3,得到y方-x方=3xy,两边同乘以-1,得到x方-y方=xy.(2x方+3xy-2y方)/(x方-2xy-y方)=[2(x方-y方)+3xy]/(

1.整理得到算式[2x-x+y]/[x^2-y^2]=[x+y]/[x+y][x-y]=1/[x-y]代入得到1/72.整理得到[1-3/(a+2)]/[(a^2-2a+1)/(a^2-4)]=[(a