Đáp án:
$A = 1$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$
$\to \dfrac{xy + yz + zx}{xyz} = 0$
$\to xy + yz + zx = 0$
$\to yz = - xy - zx$
Ta được:
$x^2 + 2yz = x^2 + yz - xy - zx = (x-y)(x - z)$
Chứng minh tương tự, ta được:
$y^2 + 2zx = (y - z)(y -x)$
$z^2 + 2xy = (z - x)(z - y)$
$\to A = \dfrac{yz}{x^2 + 2yz} + \dfrac{zx}{y^2 + 2zx} + \dfrac{xy}{z^2 + 2xy}$
$\to A = \dfrac{yz}{(x -y)(x-z)} + \dfrac{zx}{(y - z)(y - x)} +\dfrac{xy}{(z - x)(z - y)}$
$\to A = -\dfrac{xy(x - y) + yz(y - z) + zx(z -x)}{(x - y)(y - z)(z - x)}$
$\to A = -\dfrac{x^2y - xy^2 + yz(y - z) + xz^2 - x^2z}{(x - y)(y - z)(z - x)}$
$\to A = -\dfrac{x^2(y - z) - x(y^2 - z^2) + yz(y - z)}{(x - y)(y - z)(z - x)}$
$\to A = -\dfrac{(y-z)(x^2 - xy - xz + yz)}{(x - y)(y - z)(z - x)}$
$\to A = \dfrac{(x - y)(y - z)(z - x)}{(x - y)(y - z)(z - x)} = 1$
