tính giá trị biểu thức

V

vipboycodon

Có : $xy+yz+xz = 0$
=> $\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y} = 0$ (chia 2 vế cho xyz)
* $\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3} = (\dfrac{1}{x}+\dfrac{1}{y})^3-\dfrac{3}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3} $
= $(\dfrac{-1}{z})^3-\dfrac{3}{xy}(-\dfrac{1}{z})+\dfrac{1}{z^3}$ (vì $\dfrac{1}{x}+\dfrac{1}{y} = \dfrac{-1}{z}$)
<=> $\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3} = \dfrac{3}{xyz}$
$P = \dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2} = xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}) = 3$
 
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