xy-y² x² 2xy y²÷x²-2xy y² xy y²
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等于1/X*(X-Y)
原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=(1x-y-yx-y)•xyy-1=1-yx-y•xyy-1=-xyx-y.故答案是:-xyx-y.
解题思路:提取公因式进行分解解题过程:附件最终答案:略
将xy提取公因式变成:xy(x+y)=-3*6=-18
解题思路:本题主要利用完全平方公式进行因式分解即可求出结果解题过程:解:x²-6xy+9y²=(x-3y)2
对.前提是x不等于y
4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2=(4xy-x²-2xy-y²)/[xy(x+y)(x-y)]÷[(x-y)(x+2y
=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²
【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
题目是√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2如果是:√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2=(3/2)√x^3+xy/4+xy^2=(3
(xy-y^2)÷x-y/xy=y(x-y)×[xy/(x-y)]=xy²
原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y,由已知得(3x-2y)(x+y)=0,因为x+y≠0,所以3x
产生4种配子xyxyyyx有一种y有两种xy1y2xy1xy2y1y2产生的配子如上y1、y2是一个意思所以分为一组xyxyyy=1221
解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+
[(x-y)/(x+3y)]/[(x^2-y^2)/(x^2+6xy+9y^2)]-xy/(x+y)=(x-y)/(x+3y)*(x+3y)^2/(x-y)(x+y)-xy/(x+y)=(x+3y)/
原式=(x^2-y^2-x^2-2xy-y^2-2xy+2y^2-2xy)/xy=(-6xy)/xy=-6
解题思路:结合完全平方公式进行求解解题过程:答案见附件最终答案:略
x=1,y=0,z=9首先x、y、z都是个位数xyy可以写成100x+10y+y同理,zz可以写成10z+zyx写成10y+x等式重新代入以上化解后的式子,就是:100x+11y-11z=10y+x合
(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/
解题思路:先用十字相乘法、再运用平方差公式可解。解题过程:同学:另两道题目,我一时间没能解答出来,请你再检查一下原题。很抱歉,请你原谅!