已知x 2y=0x不等于0求分式2xy y的平方 x的平方-xy-搜答案-大师作文网

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z

x+2y=0x=-2y2xy+y2/x2-xy=-4y^2+y^2/4y^2+2y^2=-5/6

x+2y=0x=-2y2xy+y2/x2-xy=-4y^2+y^2/4y^2+2y^2=-5/6

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-3y=0x=3yx*2-xy+y*2/x*2+xy+y*2=(9y^2-3y^2+y^2)/(9y^2+3y^2+y^2)=7y^2/13y^2=7/13

令X/3=Y/4=Z/6=KX=3KY=4KZ=6K(X+Y-Z)/(X-Y+Z)=(3K+4K-6K)/(3K-4K+6K)=K/5K=1/5

设2分之x=3分之y=4分之z=k∴x=2k,y=3k,z=4kx2+y2+z2分之xy+yz+zx=(6k²+12k²+8k²)/(4k²+9k²+

原式=(2x+y)/(x-y)x-3y=0所以x=3y代入得原式=(6y+y)/(3y-y)=7y/2y=7/2

是不是求：5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问：是再答：已知是不是(x+3)²+|x+y+10|=

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

∵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

3x+4y=0,即x=-4y/32xy+y²/x²-xy=2y(-4y/3)+y²/(-4y/3)²-(-4y/3)y=(-5/3)y²/(28/9)

(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了

原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y，∵（x-2）2+|y+1|=0，∴x-2=0，y+1=0，即x=2，y=-1，则原式=16+16=32．

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3x+2y=0y=-1.5xx/y=x/-1.5x=-2/3y分之3x+3=3x/-1.5x+3=1

由题意得：3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy，∴C=13x2y+xy2−5xy3，故：C-A=13x2y+xy2−5xy3-（8x2y-6

由x/（x^2+x+1）=a可知x+1+1/x=1/ax+1/x=1/a-1(x+1/x)^2=(1/a-1)^2就能求出x^2+1/x^2+1欲求分式x^2/(x^4+x^2+1)求它的倒数不用再说

∵x2-y2=xy，∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3．再问：先化简2a+1/a²-1÷a²-a/a

x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup