x减y的平方分之x减y的代数式意义是什么-搜答案-智慧作业帮

x平方+y平方+8x+6y+25=0(x+4)²+(y+3)²=0∴x=-4;y=-3∴(x²-4y²)/(x²+4xy+4y²)-x/(x

（Y+1)/(Y-1)=XY+1=X(Y-1)XY-Y=2Y(X-1)=2Y=2/(X-1)再问：XY-Y=2你这一步错了，不过提醒我做法了Y+1=XY-X1+X=XY-Y1+X=Y(X-1)Y=(1

2x+5y=-34x+10y=-66x+15y=-9最后的答案是109/3

令x=y+3变成x-y=3把x-y=3带到x-y里4分之一(x-y)的平方-0.3(x-y)+0.75(x-y)的平方+10分之3(x-y=2.25-0.9+6.75+0.9=9

x=y+3∴x-y=34分之一(x-y)的平方-0.3(x-y)+0.75(x-y)的平方+10分之3(x-y）-2（x-y）+7=(0.25+0.75)(x-y)²+(-0.3+0.3-2

通分为分母是（x+y)⒉(x-y),即为2xy（x-y）/（x+y)⒉(x-y)和x（x+y）/（x+y)⒉(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-

∵x的平方-4x+4与（y-1）的绝对值互为相反数∴(x-2)²+|y-1|=0∴x-2=0y-1=0∴x=2y=1∴(x/y-y/x)÷(x+y)=(2/1-1/2)÷（2+1)=1/2乘

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

已知x,y满足x^2+y^2+5/4=2x+y,求代数式xy/(x+y)x^2+y^2+5/4=2x+yx^2-2x+1+y^2-y+1/4=0(x-1)^2+(y-1/2)^2=0x-1=0,x=1

因为X+2/X-Y=2所以X=2Y+2将X=2Y+2代入X+Y/3（X-Y）-（X-Y）/2（X+Y）=（3Y+2）/3（2Y+2-Y）-（2Y+2-Y）/2（3Y+2）接下来自己算

先分解因式,后乘法.(X^4-Y^4)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X+Y)(X^2+Y^2)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X^3+X^2Y+XY^2+

要使根式有意义,那么x-4≥0,且4-x≥0即x≥4,且x≤4,那么x=4所以[x-(x-4)/(x-3)]÷[(x²-4)/(x-3)]=(4-0)÷[(16-4)/(4-3)]=4÷12

已知x²+y²-2x-y+5/4=0所以(x-1)²+(y-1/2)²=0所以x=1,y=1/2所以xy/(x+y)=(1*1/2)/(1+1/2)=1/3如果

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

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+

x平方-y：这是一个二次二项式有两项,分别为"x平方"和"-y"最高次数为2所以该式为二次二项式

x/(x+y)+x/(x-y)=[x(x-y)+x(x+y)]/[(x+y)(x-y)]=2x^2/(x^2-y^2)由(2x)^2=16知x^2=4,由y=（-5）的平方正的平方根知y^2=25,代