作业帮xy-y-(x^2-y^2)^(1 2)=0通解
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/06 02:36:43
原式=x²-y²-(4x²y²+12xy²+9y²)=x²-y²-4x²y²-12xy²-
3xy是3x原式=(x-3y)(x+2y)+3x+y+2十字相乘x-3y2×x+2y1所以原式=(x-3y+2)(x+2y+1)
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)^
=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²
(x√x+x√y)/(xy-y^2)-[x+√(xy)+y]/(x√x-y√y)=[x(√x+√y)/[y(√x-√y)(√x+√y)]-[x+√(xy)+y]/{(√x-√y)[x+√(xy)+y]
【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
（xy-y^2)÷x-y/xy=y(x-y)×[xy/(x-y)]=xy²
(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)=1.=3(a-b)/(10ab)*(25a^2b^3)/((a+b)(a-b))=15/2*(ab^2)/(a+
[(x²+2xy+y²)/(xy-y²)][(x²-2xy+y²)/(xy+y²)]={(x+y)²/[y(x-y)]}{(x-y
(x+y)(x-y)-x(y+1)-y(y+1)=(x+y)(x-y)-(x+y)(y+1)=(x+y)(x-2y-1)
[(x-y)/(x+3y)]/[(x^2-y^2)/(x^2+6xy+9y^2)]-xy/(x+y)=(x-y)/(x+3y)*(x+3y)^2/(x-y)(x+y)-xy/(x+y)=(x+3y)/
原式=(x^2-y^2-x^2-2xy-y^2-2xy+2y^2-2xy)/xy=(-6xy)/xy=-6
xy+y平方+x-y-2=xy+x+y^2-y-2=x(y+1)+(y+1)(y-2)=(y+1)(x+y-2)
1、(-7x^y)(2x^y-3xy^3+xy)=-14x^(2y)+21x^(y+1)y^3-7x^(y+1)y;2、((x-y)^6)/((y-x)^3)/(x-y)=-(x-y)^3/(x-y)
(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/
x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²
原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x&
解(x-y)/(xy-y²)=(x-y)/y(x-y)=1/y
利用公式法逆向因式分解(x+y)(x^2-xy+y^2)=x^3+y^3(x-y)(x^2+xy+y^3)=x^3-y^3都是公式可以自己乘开来试下可以抵消的（x^2-xy+y^2)和（x^2+xy+