Đáp án: $1$

Giải thích các bước giải:

Ta có:

$\dfrac1x+\dfrac1y+\dfrac1z=0$

$\to xy+yz+zx=0$

$\to yz=-xy-xz$

$\to x^2+2yz=x^2+yz-xy-xz=(x-y)(x-z)$

Tương tự:

$y^2+2xz=(y-x)(y-z)$

$z^2+2xy=(z-x)(z-y)$

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

$\to A=\dfrac{yz}{(x-y)(x-z)}+\dfrac{xz}{(y-x)(y-z)}+\dfrac{xy}{(z-x)(z-y)}$

$\to A=-\dfrac{yz(y-z)+xz(z-x)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\to A=-\dfrac{yz(y-z)+xz^2-x^2z+x^2y-xy^2}{(x-y)(y-z)(z-x)}$

$\to A=-\dfrac{yz(y-z)+x(z^2-y^2)-(x^2z-x^2y)}{(x-y)(y-z)(z-x)}$

$\to A=-\dfrac{yz(y-z)-x(y-z)(y+z)+x^2(y-z))}{(x-y)(y-z)(z-x)}$

$\to A=-\dfrac{(y-z)(yz-x(y+z)+x^2)}{(x-y)(y-z)(z-x)}$

$\to A=\dfrac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1$