y=x除以sinx dy除以dx x=π除以6-搜答案-拍题作业网

8÷0.01=2x/3800×3=2xx=1200

你没写错吧,直接用合比定理得到:(x+y)/z=(y+z)/x=(z+x)/y=(x+y+y+z+z+x)/(x+y+z)=2=k所以k=2再问：不能用合比定理，我是初一的再答：(x+y)/z=(y+

/>2x+y=(4x-y-1)/2=(10+6x+2y)/4=(5+3x+y)/24x+2y=4x-y-1……1式3y=-1y=-1/34x+2y=5+3x+y……2式x=5-y=5+1/3=16/3

y=(2x+3)/(x-1)=[2(x-1)+2+3]/(x-1)=2+5/(x-1)x＜1,∴5/(x-1)＜0,∴y＜2或∵x＜1,x-1≠0,变换得y(x-1)=2x+3,即x=(3+y)/(y

令x=u²,dx=2udu∫(cos√x)/√xdx=∫cos(u)/u*(2udu)=2∫cos(u)du=2sin(u)+C=2sin(√x)+C

x/10=y/8=z/9=(x+y+z)/(8+9+10)=(y+z)/(8+9)所以：(x+y+z)/(8+9+10)=(y+z)/(8+9)所以：(x+y+z)/(y+z)=(8+9+10)/(8

d²y/dx²表示函数y(x)的二阶导数即d²y/dx²=d(dy/dx)/dx,楼主的理解只要把(dy/dx)在微分一下就行了导数其实也可以理解成除法的,即y

求导dy除以dx

两边对x求导y3+3xy2y'-6x=y+xy'(3xy2-x)y'=6x+y-y3y'=(6x+y-y3)/(3xy2-x)

x/3+y/4=9/10(1)x/5+y/4=7/10(2)(1)-(2)得：2x/15=1/5x=3/2代入(1)得：y/4=2/5y=8/5所以：方程组的解是：x=3/2；y=8/5

因为x+y除以z=x+z除以y=y+x除以x=k所以x+y除以z=x+z除以y=y+x除以x=k=(x+y+x+z+y+z)/(z+y+x)=2

-10x^3y÷（-5x^2）+25x^2y^2÷（-5x^2）=2xy-5y^2

dy/dx=-x^2/yydy=-x^2dx∫ydy=-∫x^2dx1/2y^2=-1/3x^3+c即：3y^2=-2x^3+C.

-0.2X/2=Y/3=Z/-7X/2=Y/33X=2YX=2/3YY/3=Z/-7-7Y=3ZZ=-7/3YX+Y+Z=(2/3Y+Y-7/3Y)/(Y+7/3Y)=(-2/3Y)/(10/3Y)=

原式=>5x+2y=200=>y=100-5x/2,且y100-5x/2x>12,由由x

设x/(a-b)=y/(b-c)=z/(c-a)=k∴x=(a-b)k,y=(b-c)k,z=(c-a)k∴x+y+z=k（a-b+b-c+c-a）=0

令x除以2=y除以3=z除以4=k则x=2k,y=3k,z=4kx+y+z除以x-2y+3z=(2k+3k+4k)除以(2k-6k+12k)=9k除以8k=8分之9

1.y^2-2xy+9=02yy'-2(xy'+y)=0y'=y/(y-x)2.x'(t)=-Asinty'(t)=Acostdy/dx=y'(t)/x'(t)=-cottd(dy/dx)/dt=(c

再问：第二题和第三题是求通解啊再答：是通解啊，不是有常数C吗

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.