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Đáp án: $\dfrac{yz}{{{x}^{2}}+2yz}+\dfrac{xz}{{{y}^{2}}+2xz}+\dfrac{xy}{{{z}^{2}}+2xy}=1$
Giải thích các bước giải:
Với $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
$\to \dfrac{yz+zx+xy}{xyz}=0$
$\to yz+zx+xy=0$
$\to yz=-zx-xy$
Khi đó:
$\,\,\,\,\,\,{{x}^{2}}+2yz$
$={{x}^{2}}+yz+yz$
$={{x}^{2}}+yz-zx-xy$
$=\left( {{x}^{2}}-zx \right)-\left( xy-yz \right)$
$=x\left( x-z \right)-y\left( x-z \right)$
$=\left( x-z \right)\left( x-y \right)$
Tương tự: $\begin{cases}y^2+2xz=\left(y-x\right)\left(y-z\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{cases}$
Vậy: $\dfrac{yz}{{{x}^{2}}+2yz}+\dfrac{xz}{{{y}^{2}}+2xz}+\dfrac{xy}{{{z}^{2}}+2xy}$
$=\dfrac{yz}{\left( x-z \right)\left( x-y \right)}+\dfrac{xz}{\left( y-x \right)\left( y-z \right)}+\dfrac{xy}{\left( z-x \right)\left( z-y \right)}$
$=\dfrac{-yz\left( y-z \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}-\dfrac{xz\left( z-x \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}-\dfrac{xy\left( x-y \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=-\dfrac{yz\left( y-z \right)+xz\left( z-x \right)+xy\left( x-y \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=-\dfrac{yz\left[ -\left( x-y \right)-\left( z-x \right) \right]+xz\left( z-x \right)+xy\left( x-y \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=-\dfrac{-yz\left( x-y \right)-yz\left( z-x \right)+xz\left( z-x \right)+xy\left( x-y \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=-\dfrac{-y\left( x-y \right)\left( z-x \right)+z\left( z-x \right)\left( x-y \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=-\dfrac{\left( x-y \right)\left( z-y \right)\left( z-x \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=\dfrac{\left( x-y \right)\left( y-z \right)\left( z-x \right)}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}$
$=1$
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