x分之x-y除以(x-x分之2xy-y的平方)-搜答案-搜狗做题网

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

y=[x/(x-2)-x/(x+2)/2x/(2-x)=4x/(x-2)(x+2)*(2-x)/2x=-2/(x+2)y的值是整数,x+2=1,-1,2,-2x=,-1,-3,0,-4

【x-y-(x-y分之4y平方)】【x+y-(x+y分之4x平方)】除以【3(x+y)-x-y分之8xy】=（x-y-x+4y)(x+y-x-4x/y)/(3x+3y-x-8x)=3y*(y-4x/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)/

x^2-x-6分解因式得(x+2)(x-3)x^2-4分解因式得(x-2)(x+2)故(x^2-x-6分之x^2-4+x-3分之x+2)=(x-2)/(x-3)+(x+2)/(x-3)=2x/(x-3

解原式=2(x-y)/(x+y)-(x+y)/3(x-y)=6(x-y)(x-y)/[3(x+y)(x-y)]-(x+y)x+y)/[3(x-y)(x+y)]=5(x-y)(x-y)/[(x+y)(x

自己想

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)

1.计算：x-y分之x+y+y-3分之2x=(x+y)/(x-y)+2x/(y-3)=((x+y)(y-3)+2x(x-y))/(x-y)(y-3)=(xy-3x+y^2-3y+2x^2-2xy)/(

(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

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+

[3x分之2-（x+y分之2）乘以[3x分之x+（y-x^2-xy）]除以x分之x-y{2/3x-2/(x+y)*[x/3x+(y-x²-xy)]}÷(x-y)/x题目错了吧?再问：{3x分

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(

解题思路：这是分式的除法，先变乘然后把分子分母因式分解，最后进行约分化成最简分式。解题过程：

（xy-x²）÷xy分之x²-2xy+y²乘x²分之x-y=-x(x-y)×xy/(x-y)²×(x-y)/x²=-y∴结果与x的值没有关系

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

解；(x/y-y/x)=(x²-y²)/xy(y分之x-x分之y)除以(x+y)=(x-y)/xy=0∴x=y