limxy(x^2-y^2) (x^2 y^2)-搜答案-搜搜做题作业网

x^2+Y^2>=0,这是肯定的啊,|xy|＜x∧2＋y∧2成立,|xy|绝对值里面大于0,成立,她小于零,也成立啊,所以去掉和不去掉,都一样,再计算时,你直接去掉就行

(2x-y)²-4(x-y)(x+2y)

(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

(x+y)(x+y)^2(-x-y)^3=－(x+y）^6

(x-y)^2*(y-x)^2*(x-y)^6=(x-y)^2*(x-y)^2*(x-y)^6=(x-y)^10再问：能详细一点吗?我,初一的再答：(x-y)^2=(y-x)^2=(x-y)^2*(x

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)^2+(x+y)(x-y)】除以2x=(x-y)*(x-y+x+y)/2x=(x-y)*2x/2x=x-y

[(-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

(x+y)ˆ2-2(x+y)(x-y)+(x-y)ˆ2=（x+y-x+y）^2=(2y)^4=4y^2如果满意请点击右上角评价点【满意】即可~再问：这个我看不懂，能不能写下来啊，谢

先一个一个的展开括号项,再同项合并就行了啊[x(x-y)-y(x-y)+(x+y)(x-y)]÷2x=[xx-xy-(xy-yy)+x(x-y)+y(x-y)]÷2x=[xx-xy-xy+yy+xx-

lim0>xy(x^2-y^2)/(x^2+y^2)^(3/2),是这个?x=rcost；y=rsint；r->0xy(x^2-y^2)/(x^2+y^2)^(3/2)=r^2sintcost*r^2

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

(2x-y)^2-4(x-y)(x+2y)

4x^2+y^2-4xy-4(x^2-xy+2xy-2y^2)展开化简吧!

=(x-y)[(3x+y)-(2x+3y)]*(x-y)=(x-y)*(x-y)(3x+y-2x-3y)=(x-y)(x-y)(x-2y)到底是化简,还是?=(x*x-2xy+y*y)(x-2y)=x

(1)(x^2/x)-y-x-y

（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/

解题思路：根据题意，由整式的运算的知识整理，分析可以求得解题过程：

解代数式得：-3x的平方y-2y的三次方因为1/2的平方=-1/2的平方,所以结果仍正确给分吧!嘿嘿!

原式=(x+y)(x-y)[(x+y)(x-y)-(x²+y²)]=(x+y)(x-y)(x²-y²-x²-y²)=-2y²(x+

因为数列{X}有界,所以设绝对值X

1.[(x-y)^2-2(y-x)]÷(x-y)=[(x-y)^2+2(x-y)]÷(x-y)=(x-y)(x-y+2)÷(x-y)=x-y+22.[(3a-7)^2-(a+5)^2]÷(4a-24)