3x=4y=7z,求x y z
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xy/x+y=-2,取倒数得1/x+1/y=-1/2①yz/y+z=3/4取倒数得1/y+1/z=4/3②zx/z+x=-3/4取倒数得1/x+1/z=-4/3③①+②+③得2(1/x+1/y+1/z
(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
三式相加:x+y+z+1/x+1/y+1/z=22/3三式相乘:xyz+y+x+1/z+z+1/x+1/y+1/xyz=28/3将1式代入2式得到xyz+22/3+1/xyz=28/3即:xyz+1/
根据题意得x-z-2=0(1)3x-6y-7=0(2)3y+3z-4=0(3)(2)+(3)×2得3x+6z-15=0x+2z-5=0(4)(4)-(1)得3z-3=0∴z=1把z=1代入(1)得x-
x/2=y/3=z/4=kx=2ky=3kz=4kx+y-z=2k+3k-4k=5k=5x=10y=15z=20
X+Y+Z-根号X-根号Y-根号Z+四分之三=(√x-1/2)^2+(√y-1/2)^2+(√z-1/2)^2=0所以√x=√y=√z=1/2xyz=1/64
1,x-3y+7z=0x-2y+4z=0y-3z=0y=3zx-6z+4z=0x=2zx:y:z=2z:3z:z=2:3:12,AB=AC,∠C=∠BAD=45°,CE=AF△AEC≌△BFA∠AFB
参考下面解题步骤:
由2x+3y-3z=0得:z-y=2x/3(2x+y-z)/(2x-y+z)=(2x-(z-y))/(2x+(z-y))将z-y=2x/3代入上式得:(2x+y-z)/(2x-y+z)=(2x-(2x
1、{x+y+z=301){3x+y-z=502){5x+4y+2z=403)1)+2)得:2x+y=404)3)-1)×2得:3x+2y=-205)4)×2-5)得:x=1006)6)代入5)得:y
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
如果是3/(x+y)=4/(x+z)=5/(y+z),设为=k,则x+y=3/k,x+z=4/k,y+z=5/k,那么x=1/k,y=2/k,z=3/k,所以xyz/[(x+y)(y+z)(z+x)]
x:y:z=2:3:4=4:6:8x+y+z=18x=4y=6z=8xyz=4x6x8=192
设单位为a,则X=4a,Y=7a,Z=8a所以4a+7a+2*8a=54a=2则X=8,Y=14,Z=16则XYZ=8*14*16=1792
(1/2)x=(1/3)y=(1/4)z=kx=2k,y=3k,z=4kx:y:z=2:3:4
x平方+y平方+2z平方-2x+4y+4z+7=0,则x²-2x+1+y²+4y+4+2z²+4z+2=0则(x-1)²+(y+2)²+2(z+1)&
x*x+y*y+2z*z-2x+4y+4z+7=0(x*x-2x+1)+(y*y+4y+4)+2(z*z+2z+1)=0(x-1)^2+(y+2)^2+2(z+1)^2=0x=1,y=-2,z=-1x
|x-z-2|+|3x-6y-7|+(3y+3z-4)^2=0绝对值、平方都大于等于0相加为0则各项均为0所以x-z-2=03x-6y-7=03y+3z-4=03式乘2得6y+6z=8加上2式3x+6
x-z-2=0,3x-6y-7=0,3y+3z-4=0,解得x=-1,y=-5/3,z=-3再问:过程再答:因为|x-z-2|+(3x-6y-7)的二次方+|3y+3z-4|=0,而|x-z-2|≥0