[Toán 8] Bai tập của mình

N

nhuquynhdat

Ta có $\dfrac{1}{x}+ \dfrac{1}{y}+\dfrac{1}{z}=0 \to (\dfrac{1}{x})^3+ (\dfrac{1}{y})^3+(\dfrac{1}{z})^3= \dfrac{3}{xyz}$

$\dfrac{yz}{x^2}+ \dfrac{xz}{y^2}+ \dfrac{xy}{z^2}$

$=\dfrac{xyz}{x^3}+ \dfrac{xyz}{y^3}+ \dfrac{xyz}{z^3}$

$= xyz.[(\dfrac{1}{x})^3+ (\dfrac{1}{y})^3+(\dfrac{1}{z})^3]= xyz. \dfrac{3}{xyz}=3$
 
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