Đáp án:
\[M = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{a}\\
\Leftrightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{{x + y + z}}\\
\Leftrightarrow \dfrac{{xy + yz + zx}}{{xyz}} = \dfrac{1}{{x + y + z}}\\
\Leftrightarrow \left( {xy + yz + zx} \right)\left( {x + y + z} \right) = xyz\\
\Leftrightarrow {x^2}y + x{y^2} + xyz + xyz + {y^2}z + y{z^2} + {x^2}z + xyz + x{z^2} = xyz\\
\Leftrightarrow {x^2}y + x{y^2} + {y^2}z + y{z^2} + {x^2}z + x{z^2} + 2xyz = 0\\
\Leftrightarrow \left( {{x^2}y + {z^2}y + 2xyz} \right) + \left( {x{y^2} + {y^2}z} \right) + \left( {x{z^2} + {x^2}z} \right) = 0\\
\Leftrightarrow y.\left( {{x^2} + 2xz + {z^2}} \right) + {y^2}\left( {x + z} \right) + xz\left( {x + z} \right) = 0\\
\Leftrightarrow y.{\left( {x + z} \right)^2} + {y^2}\left( {x + z} \right) + xz\left( {x + z} \right) = 0\\
\Leftrightarrow \left( {x + z} \right).\left[ {y.\left( {x + z} \right) + {y^2} + zx} \right] = 0\\
\Leftrightarrow \left( {x + z} \right)\left( {yx + yz + {y^2} + zx} \right) = 0\\
\Leftrightarrow \left( {x + z} \right).\left[ {\left( {xy + {y^2}} \right) + \left( {yz + zx} \right)} \right] = 0\\
\Leftrightarrow \left( {x + z} \right).\left[ {y\left( {x + y} \right) + z\left( {x + y} \right)} \right] = 0\\
\Leftrightarrow \left( {x + z} \right)\left( {x + y} \right)\left( {y + z} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + z = 0\\
x + y = 0\\
y + z = 0
\end{array} \right.\\
x + y + z = a \Rightarrow \left[ \begin{array}{l}
x = a\\
y = a\\
z = a
\end{array} \right.\\
\Rightarrow M = \left( {{x^5} - {a^5}} \right)\left( {{y^7} - {a^7}} \right)\left( {{z^9} - {a^9}} \right) = 0
\end{array}\)
Vậy \(M = 0\)
