已知x-y=y-z=3,且x²

(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

1／x＝p1／y＝q1／z＝rpq+qr+pr=1(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q

x:y=2:3=:6;9x:y:z=6:9:10则x＝6/（6+9+10）*50＝12y＝9/（6+9+10）*50＝18z＝10/（6+9+10）*50＝20xyz＝12*18*20＝4320

x/2=y/3=z/4=kx=2k,y=3k,z=4k2x+3y-4z/x+y+z=(4k+9k-16k)/(2k+3k+4k)=-1/3再问：确定？

已知x=y/3=z/5且x+y-2z不等于0那么y=3xz=5x（2x+3y-z）/（x+y-2z）=（2x+9x-5x）/（x+3x-10x）=6x/（-6x）=-1

4x-y+3z=0(1)2x+y+6z=0(2)()+(2)6x+9z=06x=-9zz/x=-2/3(1)*2-(2)8x-2y-2x-y=06x-3y=06x=3yx/y=1/2z/x=-2/3x

因为x:y:z=2:3:4所以设x=2k,y=3kz=4k因为x-y+z=36所以：2k-3k+4k=363k=36k=12所以x+y+z=2k+3k+4k=9k=9×12=108

联立消去X有5Y-20Z=0Y:z=4:1消去y有5X-15Y=0X:Z=3:1∴X:Y:Z=3:4:1

xyz＝x＋y＋z＜3z∴xy＜3由于x＜y,故xy＝2,x＝1,y＝2∴z＝3

柯西【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*（y+z＋x+z＋x+y）≥（x+y+z）^2即x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥（x+y+z）/2＝（3

【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+(

【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+4z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+4z^2]/[5(3z)^2+(

y=3-x带入2x+y=zx+3=z带入中间的式子3x+12-4x=2x+6+6x=0y=3z=3x=0Z=3

z=y-x'z=9-3=6x=x+z'x=3+6=9y=y-z'y=9-6=3xyz分别是9,3,6

(x+y)/z=(x+z)/y=(z+y)/xx,y,z等价x=y=z(x+y)(x+z)(z+x)/xyz=8

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

设(x+y)/2=(y+z)/3=(z+x)/4=kx+y=2ky+z=3kx+z=4k1式+3式2x+y+z=6k2x+3k=6k2x=3kx=3k/2代入1式得y=k/2代入3式得z=5k/2∵x

因为x:y:z=2:3:4所以y=3x/2z=2x又因为x+y+z=18,所以代入得到x+3x/2+2x=4.5x=18所以x=4所以y=6z=8

4x-3y-3z=0.1x-3y+z=0.21式-2式得3x-4z=03x=4zx:z=4:3x:z=12:91式+2式*3得7x-12y=07x=12yx:y=12:7所以x:y:z=12:7:9

①*2+②得5x=15zx=3z带入①得y=4z所以x:y:z=3:4:1