搜搜做题作业网x*x*y-x*y*y-x y=56
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原式=(x-y)(x²+2xy+y²-xy)-(x-y)(x²-2xy+y²+xy)=(x-y)(x²+xy+y²)-(x+y)(x&sup
求数学达人来解答x²/xy-x/y=(1-x)/y3x/4x+y-x-2y/4x+y=(3x-x-2y)/(4x+y)(1-y/y+x)÷x/y²-x²=(1-y)*(y
x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7
对.前提是x不等于y
x=4,y=0.5,x+y=4.5(与人家的做法一样……)(1)解题思路是以S3为基准,用S3表示出S1,S2,S4即可.在三角形BCD中有：S2/S3=DF/CF,故S2=(DF/CF)S3;同理,
=-(xy^2)^4+3(xy^2)^3+(xy^2)^2=-27-27-3=-57
xy-1+x-y=XY+X-Y-1(加法交换律）=（XY+X）－（Y＋1）＝X（Y＋1）－(Y+1)（提取公因式X）=(Y+1)(X-1)（提取公因式Y+1）这样可以么?
（x-y）[(x+y)(x+y)-xy]+(x+y)[(x-y)(x-y)+xy]=(x-y)(x+y)^2-（x-y)xy+(x+y)(x-y)^2+(x+y)xy=(x-y)(x+y)(x+y+x
这题少括号了吧.
原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1
x(x-y)-y(y-x)=12那么：化简得：（x+y）*（x—y)=12x、y是自然数,所以x1=2,y1=4x2=4,y2=2所以x+y-xy=-2
【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy
你确定没写错题目?后面的式子一约就变成前面的式子了,答案也是8/xy再问：怎么个约法?再答：y/xyyy约掉剩下1/x
（x+y)（x-y)-y^2+（x-y)^2-（6x^2y-2xy^2)÷2y=（x+y)（x-y)-y^2+（x-y)^2-2y（3x^2-xy)÷2y=（x+y)（x-y)-y^2+（x-y)^2
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
7或者-8再问：求过程^_^再答：两个等式两边相加
x²+y²=19(x+y)²=x²+y²+2xy=25xy=3(x-y)*(x-y)=(x-y)²=x²+y²-2xy=
x²y(x-y)-xy²(x-y)=xy(x-y)(x-y)=xy(x-y)²如果不懂,祝学习愉快!
(x+y)(x-y)-x(y+1)-y(y+1)=(x+y)(x-y)-(x+y)(y+1)=(x+y)(x-2y-1)
令y=kxx*x+kx*x=k*x+k*k*x*x(1-k*k+k)x^2-kx=0x((1-k*k+k)x-k)=0由上式得X=0或(1-k*k+k)x-k=0解得：k=(x-1+（或-）√（（1-