Đáp án:
$A = 3$
Giải thích các bước giải:
Ta có:
$\dfrac1x +\dfrac1y +\dfrac1z = 0$
$\to \dfrac{xy + yz + zx}{xyz}=0$
$\to xy + yz + zx = 0$
$\to \begin{cases}xy = -z(x + y)\\yz = - x(y+z)\\zx = -y(z + x)\end{cases}$
Ta được:
$A =\dfrac{yz}{x^2} +\dfrac{zx}{y^2}+\dfrac{xy}{z^2}$
$\to A =-\dfrac{x(y+z)}{x^2}-\dfrac{y(z + x)}{y^2} -\dfrac{z(x +y)}{z^2}$
$\to A = -\dfrac yx -\dfrac zx -\dfrac zy -\dfrac xy -\dfrac xz -\dfrac yz$
$\to A = -x\left(\dfrac1y +\dfrac1z\right) -y\left(\dfrac1z +\dfrac1x\right) -z\left(\dfrac1x +\dfrac1y\right)$
$\to A = -x\cdot\left(-\dfrac1x\right) - y\cdot\left(-\dfrac1y\right)-z\cdot\left(-\dfrac1z\right)$
$\to A = 1+ 1 + 1= 3$
