若圆x^ y^2-2x-4y-1=0上存在两点关于直线2ax by-2=0

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

(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4

1、4(x-y+1)+y(y-2x)=4x-4y+4+y²-2xy=y²-4y+4-2xy+4x=(y-2)²-2x(y-2)=(y-2x-2)(y-2)2、4x^4-1

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

原始=(4x²-4xy+y²+4x²-y²-4x²+4xy-4x)÷2x=(4x²-4x)÷2x=4x²÷2x-4x÷2x=2x-

（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

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

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

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

4(x-y+1)+y(y-2x)=4x-4y+4+y^2-2xy=(2-y)^2+2x(2-y)=(2-y)(2-y+2x)4x^4-13x²+9=4x^4-12x^2+9-x^2=(2x^

4x^2+y^2-4xy-4(x^2-xy+2xy-2y^2)展开化简吧!

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/

1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【（x+y)-(x-y)】/(x-y)(x+y)-2y/(x^2+y^2)-4y^

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

(2x-y)²-4(x-y)(x+y)=(4x²-4xy+y²)-4(x²-y²)=4x²-4xy+y²-4x²+4y&

1,-3再问：过程。。。再答：★（x²－2x)+（y²－4y）=5★（x－1）²+（y－2）²=1+4－5★（x－l）²=0,（y－2）²=

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