作业帮计算以圆 x^2 y^2=2x围成的区域为底而以曲面z= x^2 y^2
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/31 03:30:43
等于1/X*（X-Y）
(1)=(x-y)(x+1)(x-1)(2)=x(x^2+x+1/4x)=x(x+1/2)^2(3)=4x^2+4xy+y^2-x^2+4xy-4y^2=3x^2+8xy-3y^2=(3x-y)(x+
两个式子作差得到x=5带入得到y=23
(x+y)(x+y)^2(-x-y)^3=－(x+y）^6
原式=x²+xy-2y²+4x²-y²=5x²+xy-3y²
【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
(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
(-x-2y)*(3x-y)=-3x^2-5xy+2y^2
这样可以了但是你只有一条式子只可以做成这样#includevoidf(doublex,doubley);voidmain(){floata;floatb;printf("请输入两个数：");scanf
16(x+y)^8/9(x-y)^8=16/9*[(x+y)/(x-y)]^8哪个好理解用哪个.再问：有过程没。。。再答：这个过程其实没什么好研究的。。。[3(x-y)^2(x+y)^-2/4(x+y
=1/2x-1/(x+y)×[(x+y)/2x-(x+y)]=1/2x-1/2x+1=1/x+1=(x+1)/x
原式=(2x²+8xy-3xy-12y²)-(3x²-6xy+xy-2y²)=2x²+8xy-3xy-12y²-3x²+6xy-x
2x/(x^2-y^2)+3x/(y^2-x^2)+y/(x+y)(y-x)=2x/(x+y)(x-y)-3x/(x+y)(x-y)-y/(x+y)(x-y)=(2x-3x-y)/(x+y)(x-y)
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!
即xy=2²-(√3)=4-3=1原式=[(x+y)²-(x-y)²]/(x+y)(x-y)*[-(x²-y²)/x²y²]=(x
(x-2y)^2-4(2x+y)(y-2x)+4x(y-2)=x^2-4xy+4y^2-4(y^2-4x^2)+4xy-8x=x^2-4xy+4y^2-4y^2+16x^2+4xy-8x=17x^2-
解:{y-x=0①2x-y=6②①＋②得：y-x+2x-y=0+6x=6将X=6带入①中,得：y-6=0y=6∴｛x=6y=6
(x+2y)(x-2y)-(x+2y)-(x-2y)(2y-x)=x^2-4y^2-x-2y+(x^2-4xy+4y^2)=x^2-4y^2-x-2y+x^2-4xy+4y^2=2x^2-x-2y-4