x(x y)(x-y)-x(x y)的2次方,其中x y=1xy=二分之一-搜答案-大师作文网

xy+x+y+1因式分解

(x+1)(y+1)

x/(xy-y²)=x/[y(x-y)]y/(x²-xy)=y/[x(x-y)]∴公分母为：xy(x-y)通分后得x/(xy-y²)=x²/[xy(x-y)]y

x²/xy-x/y

求数学达人来解答x²/xy-x/y=(1-x)/y3x/4x+y-x-2y/4x+y=(3x-x-2y)/(4x+y)(1-y/y+x)÷x/y²-x²=(1-y)*(y

XY(XY-X^2)(X-Y)=X^2Y(X-Y)(X-Y)=(X-Y)(X^2Y-1)再问：答案是。-X2Y.我不知道过程是如何来的。能告诉我过程吗。谢谢。负X的平方Y

对.前提是x不等于y

(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^

3xy[2xy-x(y-2)+x-1]=3xy(2xy-xy+2x+x-1)=3xy(xy+3x-1)=3x^2y^2+9x^2y-3xy

xy-1+x-y

xy-1+x-y=XY+X-Y-1(加法交换律）=（XY+X）－（Y＋1）＝X（Y＋1）－(Y+1)（提取公因式X）=(Y+1)(X-1)（提取公因式Y+1）这样可以么?

=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²

这题少括号了吧.

因式分解x2+y-xy-x

x²＋y－xy－x＝﹙x²－xy﹚＋﹙y－x﹚＝x﹙x－y﹚－﹙x－y﹚＝﹙x－y﹚﹙x－1﹚

【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy

1-x-y+xy 因式分解

1-x-y+xy=(1-x)-(y-xy)=(1-x)-y(1-x)=(1-x)(1-y)

原式=x²y²/(x-y)^4×[-y³(x-y)³]/x³=-y^5/[x(x-y)]=-y^5/(x²-xy)

因式分解xy-x-y+1

xy-x-y+1=x(y-1)-y+1=x(y-1)-(y-1)=(y-1)(x-1)

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

xy-x-y+1因式分解

(x-1)(y-1)

已知x>y,且xy

Bxyy那么x为正数,因为负数a为任意有理数a^2等于0所以选B

原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x&

化简xy-1-x+y

解:原式=xy+y-x-1=y(x+1)-(x+1)=(x+1)(y-1)