x-1分之1除以(y分之1-x

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

(1)[1+(x-2分之1)]除以(2x-4)分之(x^2-1)=(x-1)/(x-2)*2(x-2)/(x+1)(x-1)=2/(x+1)[等于(x+1)分之2](2)[(x-y)^2+3(x^2-

2x-y+1的绝对值+（3x+2分之3y）平方=0则2x-y+1=03x+3y/2=0,则2x-y=-1①2x+y=0②①+②,得4x=-1,则x=-1/4①-②,得-2y=-1,则y=-1/2x+y

原式=(x²+x-2x)/(x+1)÷[(x-1)²/(x+1)(x-1)]=(x²-x)/(x+1)÷[(x-1)/(x+1)]=x(x-1)/(x+1)*(x+1)/

①3除以x分之2=3*x/2=3x/2②x分之3除以x^2分之6=3/x*x^2/6=x/2③x分之2x+1除以5分之6+12x=(2x+1)/x*5/(6+12x)=5/6x④x^2+3x+2分之x

自己想

1.7y分之12x除以8x的平方Y=12x/7y×1/8x²y=3/14xy²2.x平方-1分之x的平方-2x+1除以x的平方+x分之x-1=(x-1)²/(x+1)(x

(根号y分之x-根号x分之y)除以（根号y分之1-根号x分之1）=[(x-y)/根号xy]÷[（根号x-根号y)/根号xy]=根号x+根号y

规范而详细解答过程看图,明白就采纳,祝马年学习进步

1、(xy分之x²-y²)×(x-y分之x)-(y分之x)；=(x+y)(x-y)/xy×x/(x-y)-x/y=(x+y)/y-x/y=(x+y-x)/y=12、6xy²

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

（1）=x平方-y平房分之x(x+y)-y(x-y)+2xy=（x+y）*（x-y）分之（x+y）的平方=x-y分之x+y（2）=a-1分之2-a-1分之a+1=-1

[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

(x+y)/(x^4-y^4)÷1/(x^2+y^2)=(x+y)/(x^2+y^2)(x^2-y^2)÷1/(x^2+y^2)=(x+y)/(x^2-y^2)=(x+y)/(x+y)(x-y)=1/

(3x-4y)÷(2x+y)=1/2所以2x+y=2(3x-4y)=6x-8y4x=9y所以x/y=9/4

[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

你是不是觉得这种题很简单啊?再问：我对一下答案，我不确定再答：0分，我嫌麻烦再问：可以吗再答：算了，我不是要饭的2x/(2x-y)+y/(y-2x)*(x-1)²/(x+1)(x-1)*x(

现将括号里的通分x2-y2可约去最后的y分之2再问：能具体一点吗再答：原式=((x+y+x-y)/(x2-y2)÷（x的平方-y的平方分之xy）=2x/(x2-y2)×(x2-y2)/xy=y分之2只

原式=5x根号(xy)/3根号(y/x)*1/3根号(x/y)=5/9*根号(xy*x/y*x/y)=5x/9y根号(xy)

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