已知x,y,z>0求(xy 2yz) (x^2 y^2 z^2)的最大值
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![已知x,y,z>0求(xy 2yz) (x^2 y^2 z^2)的最大值](/uploads/image/f/4226510-38-0.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%3E0%E6%B1%82%28xy+2yz%29+%28x%5E2+y%5E2+z%5E2%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
(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
由x+3y-5z=0得x=5z-3y代入2x-y-3z=0中,得10z-6y-y-3z=07z=7y∴y=z代入x=5z-3y中得x=5y-3y=2y∴x=2y于是x∶y∶z=2y∶y∶y=2∶1∶1
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
由x+y-z=0,2x-y+2z=0可得:z=-3xy=-4x则3x-2y+5z/5x-3y+2z=3x+8x-15x/5x+12x-6x=-4x/11x=-4/11
x+2y-3z=0⑴2x+3y+5z=0⑵⑵-⑴得x+y-2z=0⑶⑶-⑴得y=z代入⑶得x=y=z所以(x+y+z)除以(x-y+z)=3
由题意得,x+(-2)=0,y+3=0,z=0,解得x=2,y=-3,z=0,所以,x+y+z=2+(-3)+0=-1.
x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1
方法一:特殊值法,假设x=0,y=1,z=-1x2+y2-z2分之一加x2+z2-y2分之一加y2+z2-x2分之一=0方法二:x2+y2-z2分之一=(x2+y2-(x+y))^2分之一=-1/(2
两式相减,得3x-4z=0x=4/3zx:z=4:3代入,得y=7/9zy:z=7:9
因为x+2y-z=0,7x-y-z=0两式相减,得:6x-3y=0,所以y=2x代入x+2y-z=0中,得:x+4x-z=0,那么z=5x那么(x+y+z)÷(2x-y-z)=(x+2x+5x)÷(2
1.z²-z+1/4=(z-1/2)².绝对值、根号、平方数都是非负的,而相加为0.所以都为0.即x=y,2y=z,z=1/2.所以x=y=1/4,z=1/2.2.2002x200
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
(x²-y²-z²)²-4y²z²=(x²-y²-z²)²-(2yz)²=(x²
4x-3y-3z=0①x-3y+z=0②①-②,得3x-4z=03x=4z由于z不等于0,故有x:z=4:3同理可得:①-4②,得9y-7z=09y=7zy:z=7:9
4x-3y+z=0(1)x+2y-8z=0(2)(1)-(2)×4得-11y+33z=0∴y=3z把y=3z代入(2)得x=2z把x=2z,y=3z代入x+y-z/x-y+2z得原式=(2z+3z-z
两式相加,得6X-5Z=0即X=5Z/6,即X/Z=5/6.再将X=5Z/6代入式1,得5Y+11Z/2=0得Y/Z=-11/10
X+Y-Z=6.aY+Z-X=2.bZ+X-Y=0.ca,b,c三式相加X+Y+Z=8.dd式-a式2Z=2Z=1d式-b式2X=6X=3d式-C式2Y=8Y=4
括号是什么意思?只有一半
两式相加,得3x-z=0可得z/x=5将z=5x代入1式13x-y=0得y/x=13所以x:y:z=1:13:5
x+y-7z=0①x-2y+5z=0②①-②得:3y-12z=0,即y=4z,2①+②得:3x-9z=0,即x=3z所以x+2y-z/y-2x+12z=11/10,再来题难点的.