\(\dfrac{21}{4x}+\dfrac{21}{4y}+\dfrac{21}{4z}=0\Leftrightarrow\dfrac{21}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{xy+xz+yz}{xyz}=0\Leftrightarrow xy+xz+yz=0\) \(\Rightarrow\left\{{}\begin{matrix}xy=-xz-yz\\xz=-xy-yz\\yz=-xy-xz\end{matrix}\right.\)
Ta có:
\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự ta có \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(\Rightarrow A=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-x\right)\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{z^2\left(x-y\right)-z\left(x^2-y^2\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(z^2-xz-yz+xy\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{\left(x\left(y-z\right)-z\left(y-z\right)\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-z\right)\left(y-z\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
