∫∫(4-x^2 y^2)dxdy,其中D是-搜答案-智慧作业帮

(2x-y)²-4(x-y)(x+2y)

写清楚点点比如说前面是X/X还是X/(X-Y)打上括号分段这样才能清楚的做

(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

(2x+y)(2x-3y)+x(2y+y)=(2y+y)(2x-3y+x)=(2y+y)(3x-3y)=3(2y+y)(x-y)2a(x-y)-4b(x-y)=2(x-y)(a-2b)

(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

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

x=rcosθy=rsinθ∫∫(D)arctany/xdxdy=∫∫(D')arctan(sinθ/cosθ)rdrdθ其中D':1

(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!

4x(y-x)-y^2因式分解

4x(y-x)-y²=4xy-4x²-y²=-(4x²-4xy+y²)=-(2x-y)²很高兴为您解答,【the1990】团队为您答题.请点

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

(2x-y)(2x+y)-2(3x-4y)(4y+3x)=4x²-y²-2(9x²-16y²)=4x²-y²-18x²+32y&s

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

因为√x(√x+2√y)=√y(6√x+5√y),所以x+2√(xy)=6√(xy)+5y,所以x-4√(xy)-5y=0,所以(√x+√y)(√x-5√y)=0,所以√x+√y=0或√x-5√y=0

直接用常规积分解比较繁琐,而且涉及到特殊形式积分,改为（r,θ)坐标,即∫∫4r^2drdθ,其中θ积分限为（0,2π）,r为（0,1）,这样积分得8/3πr^3|（0,1）,结果为8/3π

(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-

“4y^2/x-y”应该是“4y^2/x-2y”吧x+2y+4y^2/x-2y+4x^2y/4y^2-x^2=(x+2y)+4y^2/(x-2y)+4x^2y/(2y-x)(2y+x)=(x+2y)+

x=rcost,y=rsint,代入方程得r^2

x^2+y^2=x+y化成标准式(x-1/2)^2+(y-1/2)^2=1/2x=1/2+rcosαy=1/2+rsinαα∈[0,2π]r∈[0,√2/2]∫∫(x+y)dxdy=∫∫(1+rcos

(x+3y-4)(2x-y)

原式等于2x^2-xy+6xy-3y^2-8x+4y=2x^2+5xy-3y^2-8x+4y

x+2y+4y^2/(x-2y)=((x+2y)(x-2y))/(x-2y)+4y^2/(x-2y)=(x^2-4y^2)/(x-2y)+4y^2/(x-2y)=x^2/(x-2y)或者是你的题说的有