且xyz=0求x:y:z
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 03:22:06
![且xyz=0求x:y:z](/uploads/image/f/1320093-45-3.jpg?t=%E4%B8%94xyz%3D0%E6%B1%82x%3Ay%3Az)
(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
设(y+z)/x=(x+z)/y=(x+y)/z=k;y+z=kx;x+z=ky;y+z=kx;2(x+y+z)=k(x+y+z);k=2或x+y+z=0;所以,(y+z)(x+z)(x+y)/xyz
是X+Y/5=Y+X/6=Z+X/7吧由X+Y/5=Y+X/6解得,X=24Y/25把上式代入:Y+X/6=Z+X/7解得Z=179Y/175所以X:Y:Z=(24Y/25):Y:179Y/175=1
设(y+z)/x=(z+x)/y=(y+x)/z=k则y+z=kx,z+x=ky,y+x=kz三式相加2(x+y+z)=k(x+y+z)故当x+y+z=0时,k=-1,但xy-z不等于0,可知x+y+
令(y+z)/x=(z+x)/y=(x+y)/z=t∴y+z=xt,z+x=yt,x+y=zt三式相加得:2(x+y+z)=(x+y+z)t∴(2-t)(x+y+z)=0∴2-t=0或x+y+z=0若
x+y+z=0所以x+y=-zx+z=-yy+z=-xx(1/y+1/z)+y(1/x+1/z)+z(1/x+1/y)=x/y+x/z+y/x+y/z+z/x+z/y=(x+y)/z+(x+z)/y+
∵xyz(x+y+z)=1.∴x(x+y+z)=1/(yz)即x²+xy+xz=1/(yz)∴(x+y)(x+z)=x²+xz+xy+yz=(yz)+[1/(yz)]≥2当且仅当y
2x-3y+z=0①3x-2y-6z=0②由①×6得12x-18y+6z=0③由②+③得15x-20y=015x=20yx=4y/3把x=4y/3代入①得8y/3-3y+z=0-y/3+z=0z=y/
4x-5y+2z=0①x+4y-3z=0②由①+②得:5x-(y+z)=05x=y+z由②*4-①得:21y-14z=0y=2z/35x=2z/3+z=5z/3x=z/3x:y=(z/3):(2z/3
因为xyz不等于0,则根据2x-y+z=0有2x=y-z.又x+2y-5z=0有x=5z-2y.所以就有2(5z-2y)=y-z算得y=11/z,x=3/5z然后分别把x.y用含有z的式子代入(x-y
2x-y+z=0上式两边同时除以x可得y/x-z/x=2标注为①x-2y+3z=0上式两边同时除以x刻碟2y/x-3z/x=1标注为②联解①与②可得y/x=5,z/x=3所以(x²+3y
因为|a|/a不是等于1就是-1,故|X|/X+|Y|/Y+|Z|/Z=1代表其中XYZ中有两个大于0,一个小于0故XYZ/|XYZ|=-1
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
(x+y)/z=(x+z)/y=(z+y)/xx,y,z等价x=y=z(x+y)(x+z)(z+x)/xyz=8
根据题意得20(x+y)=12(y+z)12(y+z)=15(z+x)化简得5x+2y=3z ①5x-4y=-z ② ①-②得6y=4z即y=2/3z①*2+②得15x=5z 即 x=1/3zx:
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
(y+z)/x=(z+x)/y=(x+y)/z=ty+z=tx①z+x=ty②x+y=tz③①+②+③得(x+y+z)*2=t(x+y+z)t=2①*②*③得(y+z)(z+x)(x+y)/(xyz)
x:y:z=2:3:4=4:6:8x+y+z=18x=4y=6z=8xyz=4x6x8=192
(1/2)x=(1/3)y=(1/4)z=kx=2k,y=3k,z=4kx:y:z=2:3:4