若x>0 y>0 x分之2 +y分之8 等于1

x+y+3=0,x+y=-33分之2-2x-2y+（-x-y）的平方=【2-2（x+y)+（x+y)^2】/3=（2-6+9）/3=5/3（x+y）/xy=3,x+y=3xy(2x-3xy+2y)/(

4x-5y=04x=5yx=(5/4)y原式=[(5/4)y-y]/[(5/4)y+y]=(1/4)y/(9/4)y=1/9

是3x+2y-x分之x-y-z吗?还是3x+2y-z分之x-y-z设x/2=y/3=z/4=K则x=2k,y=3k,z=4kx-y-z/3x+2y-z=2k-3k-4k/6k+6k-4k=-5k/8k

2x-y+1的绝对值+（3x+2分之3y）平方=0则2x-y+1=03x+3y/2=0,则2x-y=-1①2x+y=0②①+②,得4x=-1,则x=-1/4①-②,得-2y=-1,则y=-1/2x+y

|2分之5x-y|+y²-10y+25=0|2分之5x-y|+(y-5)²=0∴5x/2-y=0y-5=0∴x=2y=51-x-2y分之x-y×x平方-y平方分之x²-4

令2分之x=3分之y=4分之z=k,则x=2k,y=3k,z=4k,(x+3y-2z)/(2x+y)=(2k+9k-8k)/(4k+3k)=(3k)/(7k)=3/7

解由5分之x=4分之y=3分之z令5分之x=4分之y=3分之z=k则x=5k,y=4k,z=3kk不等于0则3x-2y分之x+y+z=（3x-2y）分之（x+y+z）=（3*5k-2*4k）分之（5k

②*2-①得,5/y+11=0∴y=-5/11代入②得,1/x-44/5+6=0∴x=5/14经检验,符合题意∴方程组的解是：x=5/14y=-5/11

令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0

x平方+y平方-4x-2y+5=0(x-2)^2+(y-1)^2=0(x-2)^2=0(y-1)^2=0x=2y=1根号8y-2x分之（根号x）+y=(根号2)+1/根号(8*1-2*2)=[(根号2

令2分之X=3分之y=4分之z=mX+Y+Z=2m+3m+4m=9mX+Y-Z=2m+3m-4m=m一比等于9

用反证法：假设1/x+y=2;因为x>0y>0,所以1+y≥2x1+x≥2y,所以1+y+1+x≥2x+2y即2≥x+y,与条件x+y>2相矛盾即假设不成立故原命题成立

1.令x/2=y/3=z/4=k,则有:x=2ky=3kz=4k所以:(2x+y-z)/(x+y+z)=(2×2k+3k-4k)/(2k+3k+4k)=1/32因为x^2+y^2+2x+4y+4=0即

3x-5y=03x=5yx/y=5/3(x-2y)/(x+3y)=(x/y-2y/y)/(x/y+3y/y)=(x/y-2)/(x/y+3)=(5/3-2)/(5/3+3)=(-1/3)/(14/3)

1.若x分之1-y分之1=2,则求x-xy-y分之3x+xy-3y=0的值.若x分之1-y分之1=2变形（y-x）/xy=2,(y-x)=2xy(3x+xy-3y)/(x-xy-y)=[3(x-y)+

再答：不明白的地方可以追问，望采纳，谢谢

3X=4Y两边除以3YX/Y=4/3(X-Y)/X=X/X-X/Y=1-4/3=-1/3

∵(1+X)²+Y+1的绝对值=0∴1+x=0y+1=0x=-1y=-1∴4x-3y分之2x-y=(-2+1)/(-4+3)=(-1)/(-1)=1

5x=-3yx=-3|5y直接带进去得答案,或者=2\3x-2|3x+2|x+y的和乘（x+y）除以x-y|x=2乘x\x-y=2（-3\5y）\（-3\5y-y）=3\4手机果然不行.

解；(x/y-y/x)=(x²-y²)/xy(y分之x-x分之y)除以(x+y)=(x-y)/xy=0∴x=y