求z=x²y(5-x-y)在闭区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/21 11:00:32
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
5x+3y=3z--------a-x-3y=-z--------ba式+b式4x=2z得z=2x代入a中得x=3y,y=x/3x:y:z=x:x/3:2x=1:1/3:2=3:1:6
x:y:z=(3y/5):y:(7y/4)=(3/5):1:(7/4)=12:20:35再问:已知x+2y-z=02x+3y+z=0求x:y
x=4y/3y=yz=2y/5所以,x:y:z=4/3:1:2/5=20:15:6
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(5x+3y+2z)+(4x+6y+7z)=2011+20129(x+y+z)=4023x+y+z=447
x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y
设x/3=y/4=z/5=k,则x=3ky=4kz=5k带入x+y+z/3x-2y+z,最后约去k就可以了
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,
设x/3=y/4=z/5=m则x=3m,y=4m,z=5m则x+y+z/3x-2y+z=(3m+4m+5m)/(9m-8m+5m)=12/6=2
∵y+z÷x=Z+X÷y=X+Y÷z容易发现x,y,z位置互换也成立∴式子与x,y,z值无关∴x=y=z∴(X+Y-Z)÷(X+Y+z)=x/3x=1/3明教为您解答,请点击[满意答案];如若您有不满
因为x:y:z=3:4:5所以设x=3k,y=4k,z=5k(k≠0)(1)z/(x+y)=5k/(3k+4k)=5k/7k=5/7(2)x+y+z=63k+4k+5k=612k=6k=1/2x=3k
设:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y=k{x+y-z=kz(1){y+z-x=kx(2){z+x-y=ky(3)(1)+(2)+(3)得:(x+y+z)=k(x+y+z)(x
依题意得,3y=4x,所以y=4/3x,同理,z=5/4y=5/3x,代入得1
令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+5z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(
设x+y-z/z=x-y+z/y=y+z-x/x=k有x+y-z=kzx-y+z=kyy+z-x=kx三式相加得x+y+z=k(x+y+z)k=1得x+y=(k+1)zx+z=(k+1)yy+z=(k
x:y:z=20:15:63x=4y,x:y=4:3=20:152y=5z,y:z=5:2=15:6x:y:z=20:15:6
应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11
解:不防设x=2A,则y=3A,z=5A.由x+y+z=20,可知2A+3A+5A=20,10A=20,A=2.则x=4,y=6,z=10.