Question

xy/(2y+4x) = yz/(4z+6y) = xz/(6x+2z) và x^2 + y^2 + z^2 = 28

Tìm x,y,z

Answer 1

`(xy)/(2y+4x)=(yz)/(4z+6y)=(xz)/(6x+2z)`

`=>(xy)/(y+2x)=(yz)/(2z+3y)=(xz)/(3x+z)`

`=>(y+2x)/(xy)=(2z+3y)/(yz)=(3x+z)/(xz)`

`=>1/x+2/y=2/y+3/z=3/z+1/x`

Đặt: `1/x=a;1/y=b;1/z=c`

Ta có: `a+2b=2b+3c=3c+a`

`a+2b=2b+3c=>a=3c`

`2b+3c=3c+a=>a=2b`

`=>a=2b=3c=>1/x=2/y=3/z`

`=>x/1=y/2=z/3`

Lại đặt: `x/1=y/2=z/3=k`

`=>x=k;y=2k;z=3k`

`x^2+y^2+z^2=k^2+4k^2+9k^2=14k^2=28`

`<=>k^2=2`

Với `k=\sqrt{2}=>x=\sqrt{2};y=2\sqrt{2};z=3\sqrt{2}`

Với `k=-\sqrt{2}=>x=-\sqrt{2};y=-2\sqrt{2};z=-3\sqrt{2}`