dx dy=x^3y^3 xy的通解

x-y=2xy3x-5xy-3y/x+xy-y=[3(x-y)-5xy]/[(x-y)+xy]=(6xy-5xy)/(2xy+xy)=1/31/x+1/y=4y+x=4xyx-5xy+y/2x+xy+

T1＜T2首先T1=∫∫(x+y)^2dxdyT2=∫∫(x+y)^3dxdy.这两个相除（x+y).你仔细想一下,如果（x+y）始终＞=1,或者始终＜=1,那么就好判断了.因此现在问题就看在D范围内

xy/x+y=3xy=3(x+y)3x-5xy+3y/-x+3xy-y=(3x+3y)-5*3(x+y)/[-x-y+3*3(x+y)]=-12(x+y)/8(x+y)=-3/2望采纳,谢谢!

y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5

题目条件中少写了一点：上半球面取上侧由积分曲面方程知：x²+y²+z²=a²则分母化为a²变成常数提出；补平面Σ1：z=0,x²+y

3x=xy+3y,xy=3x-3y2x+xy-2y/y-x-xy=(2x+3x-3y-2y)/(y-x-3x+3y)=(5x-5y)/[-(2x-2y)]=5/(-2)=-5/2

积分区域：0≤x≤1,0≤y≤x∫∫3xy^2dxdy=3∫xdx∫y^2dy=3∫x[y^3/3]dx=3∫x*x^3/3dx=∫x^4dx=x^5/5=1/5

原式=(3x+3y-xy)/[3xy-(x+y)]=[3(x+y)-xy]/[3xy-(x+y)]=[3-xy/(x+y)]/[3xy/(x+y)-1]=(3-2)/(3×2-1)=1/5

因为xy/x+y=3,所以xy=3（x+y）（1）将式子（1）代入求值式子：2x-3xy+2y/-x+3xy-y=2x-9x-9y+2y/-x+9x+9y-y=-7x-7y/8x+8y=-7/8

∫∫xy²dxdy=∫dθ∫(rcosθ)*(rsinθ)²*rdr(应用极坐标变换)=∫(cosθsin²θ)dθ∫r^4dr=∫sin²θd(sinθ)∫r

xy/x+y=3即：xy=3(x+y)(2x-3xy+2y)/(-x+3xy-y)=[2(x+y)-3xy]/[3xy-(x+y)]=[2(x+y)-9(x+y)]/[9(x+y)-(x+y)]=-7

xy/x+y=6∴xy=6(x+y)3x-2xy+3y/-x+3xy-y=[3x-12(x+y)+3y]/[-x+18(x+y)-y]=-9(x+y)/17(x+y)=-9/17再问：你在啊？再答：在

解此题,应先大致画出图形后去求解.因为不能画图和写公式,我只能写出答案为ln3.

xy/x+y=3xy=3（x+y）3x-5xy+3y/-x+3xy-y=3x-15（x+y）+3y/-x+9（x+y）-y=-12(x+y)/8（x+y）=-3/2

16524

3(x^2-xy)-xy+y^2=3(x^2-xy)-(xy-y^2)=3*5-(-3)=15+3=18

∫∫√(y²-xy)dxdy=∫dy∫√(y²-xy)dx=∫dy∫√(y²-xy)(-1/y)d(y²-xy)=∫{(-1/y)(2/3)[(y²-

答案见图

令x=x^2,得到x=0和x=1,所以积分区域x是在0到1之间,而且在此区域里,x>x^2显然不能直接对(sinx/x)dx进行积分,所以先对dy进行积分∫∫（sinx/x）dxdy=∫(上限1,下限

∫∫(D)(x²+y)dxdy=∫(1→2)dx∫(1/x→x)(x²+y)dy=∫(1→2)[x²y+y²/2]|(1/x→x)dx=∫(1→2)[x