BĐT đại số

H

hocgioi2013

Ta có 1/x+1/y+1/z=0
=>1/x+1/y=-1/z
=>(1/x+1/y)^3= (-1/z)^3
=>1/x^3+1/y^3+3.1/x.1/y.(1/x+1/y) =-1/z^3
=>1/x^3+1/y^3+1/z^3= -3.1/x.1/y.(1/x+1/y) =3/(xyz) (vì 1/x+1/y=-1/z)
Mặt khác: 1/x+1/y+1/z=0
=>(xy+yz+zx)/(xyz)=0
=>xy+yz+zx=0
A=yz/x^2 +2yz + xz/y^2+ 2xz + xy/z^2+ 2 xy
=xyz/x^3+xyz/y^3+xyz/z^3 +2(xy+yz+zx) (vì x,y,z khác 0)
=xyz(1/x^3+1/y^3+1/z^3) (vì xy+yz+zx=0)
=xyz.3/(xyz) (vì 1/x^3+1/y^3+1/z^3=3/(xyz) )
=3
Vậy A=3

p/s:chonhoi110 mình làm biến gõ latex quá sửa lại giúp mình nha thanks =))
 
H

hiendang241

ádfghjk

ta có $\frac{1}{x}$+$\frac{1}{y}$+$\frac{1}{z}$=0
nên xy+yz+zx=0\Rightarrow xy=-xz-yz;xz=-xy-yz;yz=-xy-yz
\Rightarrow $\frac{xy}{z^2+2xy}$ +$\frac{zy}{x^2+2yz}$+$\frac{xz}{y^2+2xz}$
=$\frac{xy}{z^2+xy-xz-yz}$+$\frac{yz}{x^2+yz-xy-xz}$+$\frac{xz}{y^2+xz-xy-yz}$
=$\frac{xy}{(y-z)(x-z)}$+$\frac{yz}{(x-y)(x-z)}$+$\frac{xz}{(z-y)(x-y)}$
=$\frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}$
=1
 
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