(2x(Y-1)^5-3x^2(y-1)^4

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

（x+2)(x+3)=x²+5x+6(x-4)(x+1)=x²-3x-4(y+4)(y-2)=y²+2y-8(y-5)(y-3)=y²-8y+15x²

方程两边同时进行通分可得[5(x2-16)+96]/(x2-16)=[(2x-1)(x-4)+(3x-1)(x+4)]/(x2-16)再问：不好意思这是两道题再答：这好似第一个原式=(x+y)[(x-

[(x+2y)(x-2y)-(2x-y)^2+(3x-y))2x-5y)]/(-x)=(x²-4y²)-(4x²-4xy+y²)+(6x²-17xy+

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)（2a+b)(2

因为(x-y)/(x+y)=3,则(x＋y)/(x－y)＝1/3则5(x-y)(x+y)-(x+y)/2(x-y)＝5*3-1/(3*2)=15-1/6=89/6

(x-y)^5*(y-x)^3*(x-y)^2=-(x-y)^5*(x-y)^3*(x-y)^2=-（x-y）^(5+3+2)=-(x-y)^10

（2x+(y-1)）(2x-(y-1))=4x^2-(y-1)^2(3x+y)(3x-y)=9x^2-y^2相减得4x^2-y^2+2y-1-9x^2+y^2=-5x^2+2y-1带入得-5*1/25

(4x-2y)-[5x-(8y-2x-y)]+x=4x-2y-5x+8y-2x-y+x=-2x+6y-(2x²-3x-1)-2(x²-3x+5)+(x²+4x+3)=-2

1)(2x+5y)(2x-5y)(-4x^2-25y^2)=-(4x^2-25y^2)(4x^2+25y^2)=-(16x^4-625y^4)=625y^4-16x^42)[（x+y)(x-y)-(x

（1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

13y-4x

x（x*x-1)+x(5-x*x)-6y+7=-x+5x-6y+7=2(2x-3y)+7=2*(-4)+7=-8+7=-1再问：能在写详一点吗-x+5x-6y+7再答：x（x*x-1)+x(5-x*x

(2X-3y)-(5x-y)=2x-3y-5x+y=2x-5x-3y+y=(2x-5x)+(-3y+y)=-3x-2y这样写总是全过程了吧

原式=[(5x²+8xy-4y²)-(9x²-4y²)]÷4x=(-4x²+8xy)÷4x=-x+2y当x=2,y=-1时原式=-2-2=-4

解3(x-1)-(x-5)=3x-3-x+5=(3x-x)+(5-3)=2x+22(x²-y²)-3(x²-2y²)=(2x²-3x²)+(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009）y=0+(1×1004-2009)y=-1005y=1005/2