2y分之x括号平方乘2x分之y

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原式=[(x+2y)/(x+y)][xy/(x+2y)]÷[(x+y)/xy]=[xy/(x+y)]×xy/(x+y)=x²y²/(x+y)²

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

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（x的平方-y的平方)分之2xy+(x+y)分之x-(y-x)分之y=（x的平方-y的平方)分之2xy+(x+y)分之x+(x-y)分之y=(x²-y²)分之[2xy+x(x-y)

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xy分之x的平方-2xy+y的平方乘x的平方分之x-y=x(y-x)*xy/(x-y)²*（x-y）/x²=-y再问：不好意思，打错了(xy-y的平方）除以x

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

（x分之x-y）-（y分之x+y）+（2xy分之x的平方-y的平方）=1-y/x-1-x/y+(x^2-y^2)/2xy=-2(y^2+x^2)/2xy+(x^2-y^2)/2xy=-(x^2+3y^

∵x^2+y^2-2x+4y+5=0配方：∴(x-1)^2+(y+2)^2=0那么x-1=0且y+2=0所以x=1,y=-2∴(x^4-y^4)/(2x^2+xy-y^2)*（2x-y)/(xy-y^

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)

请把题目说清楚这很容易产生歧义

y=2x/3则xy=2x2/3y2=4x2/9所以原式=(2x2/3)/(x2+4x2/9)-(4x2/9)/(x2-4x2/9)=(2/3)/(1+4/9)-(4/9)/(1-4/9)=6/13+4

=(x+y)(x-y)/(x+y-1)(x-y+1)×(x-y+1)(x-y-1)/(x+y)²×(x+2y)(x-y)/x(x-y)=(x-y-1)(x+2y)(x-y)/x(x+y-1)

x+2y+4y^2/(x-2y)-4x^2y/(x-4y^2)=x+2y+4y^2/(x-2y)-4x^2y/[(x+2y)(x-2y)]通分,分母为[(x+2y)(x-2y)]=(x+2y)(x+2

原式=[(x+2y)/(x+y)]×[xy/(x+2y)]÷[(x+y)/xy]=[(x+2y)/(x+y)]×[xy/(x+2y)]×[xy/(x+y)]=x²y²(x+2y)/

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+

把4拆成3+1(x²-2√3x+3)+(y²+2y+1)=0(x-√3)²+(y+1)²=0平方大于等于0相加等于0,若有一个大于0,则另一个小于0,不成立.所

是这样的形式吗?[x/（x+y）+2y/（x+y）]×[xy/（x+y）]÷（1/x+1/y）一:原式=（x+2y）/（x+y）×[xy/（x+2y）]÷[（x+y）/xy]=x²y&sup

(3y/2x)×(4x²/9y²)=2x/3y;如果本题有什么不明白可以追问,