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Với x,y,z khác 0 ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{yz+xz+xy}{xyz}=0=>yz+xz+xy=0\)
Ta luôn có nếu a+b+c=0 thì a3+b3+c3=3abc
Vì xy+yz+zx=0 nên x3y3+y3z3+z3x3=3x2y2z2
Với x3y3+y3z3+z3x3=3x2y2z2 ta có:
\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
Vậy ....
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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.
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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)