(2y分之x)的平方乘以2x分之y减去y的平方-搜答案-百度作业网

由x-3y=0可知x=3y原式=(2x+y)(x-y)/(x-y)²=(2x+y)/(x-y)=(6y+y)/(3y-y)=7y/2y=7/2

[3a^3/(x+y)]^3*(x^2-y^2)/[(y-x)/(x+y)]^2=27a^9/(x+y)^3*(x+y)*(x-y)*(x+y)^2/(x-y)^2=27a^9/(x-y)

(3x^2/4y)*(2y/3x)+(x^2/2y^2)/2y^2/x=x/2+x^3/4y^4=(2y^4*x+x^3)/4y^4=x(x^2+2y^4)/4y^4.

(x分之x的平方-y的平方)乘以(x的平方-2xy+y的平方分之2x)其中x=2y=1=(x^2-y^2)/x*2x/(x-y)^2=(x+y)(x-y)/x*2x/(x-y)^2=2(x+y)/(x

∵y-2x=0∴y=2x∴原式＝x^2+x(2x)+x^2(2x)^2-x^2/2x-(2x)^2＝x^2+2x^2+5x^2-x/2-4x^2＝4x^2-x/2

x减1分之x平方减1,化简为x+1得2005+1=2006第二问不太清楚

1）3x2）2x分之3y3）x分之x-y求采纳,以后有问题直接问我再问：求过程。谢谢好人~再答：这也要过程……？！好吧我试试1）原式=根号下{（2x分之y）×（2y分之3x²）×（x分之12

x-3y=0x=3y原式=(2x+y)(x-y)/(x-y)²=(2x+y)/(x-y)=(6y+y)/(3y-y)=7y/2y=7/2

=(x+2y)(x-2y)/(x+2)²×(x+2)/3x(x+2y)=(x-2y)/3x(x+2)=(x-2y)/(3x²+6x)

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

没写完,补充,向我追问再问：当x的二次方+4y的二次方-4x+4y+5=0时2x的二次方+xy-y的二次方分之x的四次方-y的四次方乘以xy-y二次方分之2x-y乘以（x二次方+y二次方）的平方分之y

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

（x/2y）²×（y/2x）-x/y²÷（2y²/x）=x²/2y×（y/2x）-（x/y²）×（x/2y²）=x/8y-x²/（

X的平方+Y的平方-6X-8Y+25=0,整理就是x²+y²-6x-8y+9+16=0即x²-6x+9+y²-8y+16=0于是（x-3)²+（y-4

(x+1)分之1-(x²-1)分之1×(x+1)分之(x²-2x+1)=(x+1)分之1-[(x+1)(x-1)]分之1×(x+1)分之(x-1)²=(x+1)分之1-(

[1]y+2分之y的平方-6y+9除以[3-y]=（3-y)²/(y+2)×/(3-y)=(3-y)/(y+2)[2]x的立方分之x的平方y乘以[-y分之1]=x²y/x³

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