x减2y分之x减y除以x的平方减4xy加4y分之x-搜答案-搜狗做题网

因为x=1+根号2y=1减根号2所以xy=1+根号2乘以1减根号2=1-2=-1x+y=2所以可以求得,原式=xy/(x+y)=-1/2

1-（x-y）/(x+2y)÷(x²-y²)/(x+2y)²=1-(x-y)/(x+2y)×(x+2y)²/(x+y)(x-y)=1-(x+2y)/(x+y)=

3x^2+xy-2y^2=(3x-2y)(x+y)=0x=-2y/3,或x=-y但由于你要求的式子中分母里有X平方-Y平方x=-y舍去其他就代入吧

x^2=x的平方‘/’为除以[(x+2y)^2-(x+y)(x-y)-5y^2]/(2x)=[(x^2+4xy+4y^2)-(x^2-y^2)-5y^2]/(2x)=[4xy]/(2x)=2y[(3x

前者分子：x^2-y^2=(x+y)(x-y)那前面这项就可以化简约分啦后者分子：4x^2-4xy+y^2=(2x)^2-2*2xy+y^2=(2x-y)^2建议你理解一下x^2-2xy+y^=(x-

=(xy-x的平方）除以(x的平方-2xy+y的平方\xy)乘以(x-y\x的平方)=-x（x-y）×xy/（x-y）2×（x-y）/x2=-y再问：嗯

﹙x²－4y²﹚/﹙x²＋2xy＋y²﹚÷﹙x＋2y﹚/﹙2x²＋2xy﹚＝﹙x－2y﹚﹙x＋2y﹚/﹙x＋y﹚²·2x﹙x＋y﹚/﹙x＋2

x平方+y平方+2xy分之x平方-4y平方除以x平方+xy分之2y+x=(x-2y)(x+2y)/(x+y)²·x(x+y)/(x+2y)=x(x-2y)/(x+y)

原式=[2x/(x-y)]*[y/(x+y)]-[x^4/(x^4-y^4)]/[x^2/(x^2+y^2)]^2=2xy/(x^2-y^2)-(x^2+y^2)^2/[(x^2+y^2)(x^2-y

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x(y-x)÷(x²-2xy+y²)/xy×(x-y)/x²=-x(x-y)×xy/(x-y)²×(x-y)/x²=-y

(x^2-4y^)\(x+y)^2乘以2x(x+y)/(x+2y)=2x(x-2y)\(x+y)

解题思路：把分式的分子分母分解因式再根据法则进行计算，能约分的要先约分解题过程：答案见附件

[1/(x-y)-1/(x+y)]/[xy^2/(x^2-y^2)]=[(x+y-x+y)/(x-y)(x+y)]/[xy^2/(x-y)(x+y)]=[2y/(x-y)(x+y)]/[xy^2/(x

拍张图片再问：再答：再答：有数学题可以求助我

原式=[(x+y)/(x-y)]^2*[2(x-y)/5(x+y)-[X/(x^2y^2)]/[x/(x+y)]=2/5-[x/(x^2y^2)]*[(X+y)/x]=2/5-(x^2y^2)/(x+

原式=[x/(x-y)]*[y^2/(x+y)]-[x^4y/(x^2+y^2)(x^2-y^2)]/[x^2/(x^2+y^2)]=xy^2/(x^2-y^2)-x^2y/(x^2-y^2)=xy(

能把等式照下来吗

【（x+2y）2-(x+y)(3x-y)-5y2】÷（2x）=[x^2+4xy+y^2-(3x^2+2xy-y^2)-5y^2]/(2x)=[-2x^2+2xy]/(2x)=y-x=1/2-2=-3/

(x+y)分之(x²-y²)×(x²+xy)分之(2x+2y)÷(x-y)=(x+y)分之[(x+y)(x-y)]×[x(x+y)]分之[2(x+y)]×(x-y)分之1