(0,5 điểm) Cho \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{{x + y + z}}.\) Chứng minh rằng:

\[\frac{1}{{{x^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{{x^{2023}} + {y^{2023}} + {z^{2023}}}}.\]

Hướng dẫn giải

Theo giả thiết, \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{{x + y + z}}.\)

Suy ra \(\frac{{yz + xz + xy}}{{xyz}} = \frac{1}{{x + y + z}}.\)

\(\left( {yz + xz + xy} \right)\left( {x + y + z} \right) = xyz\)

\(yz\left( {x + y + z} \right) + xz\left( {x + y + z} \right) + xy\left( {x + y + z} \right) = xyz\)

\[xyz + {y^2}z + y{z^2} + {x^2}z + xyz + x{z^2} + {x^2}y + x{y^2} + xyz = xyz\]

\[\left( {{x^2}z + 2xyz + {y^2}z} \right) + \left( {y{z^2} + x{z^2}} \right) + \left( {{x^2}y + x{y^2}} \right) = 0\]

\[z{\left( {x + y} \right)^2} + {z^2}\left( {x + y} \right) + xy\left( {x + y} \right) = 0\]

\[\left( {x + y} \right)\left[ {z\left( {x + y} \right) + {z^2} + xy} \right] = 0\]

\[\left( {x + y} \right)\left( {xz + yz + {z^2} + xy} \right) = 0\]

\[\left( {x + y} \right)\left[ {\left( {xz + xy} \right) + \left( {yz + {z^2}} \right)} \right] = 0\]

\[\left( {x + y} \right)\left[ {x\left( {y + z} \right) + z\left( {y + z} \right)} \right] = 0\]

\[\left( {x + y} \right)\left( {y + z} \right)\left( {x + z} \right) = 0\]

Suy ra \(x + y = 0\) hoặc \(y + z = 0\) hoặc \(x + z = 0.\)

⦁ Nếu \(x + y = 0\) thì \(x =  - y,\) khi đó \[{x^{2023}} =  - {y^{2023}}.\]

Ta có \[\frac{1}{{{x^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{ - {y^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{{z^{2023}}}};\]

\[\frac{1}{{{x^{2023}} + {y^{2023}} + {z^{2023}}}} = \frac{1}{{ - {y^{2023}} + {y^{2023}} + {z^{2023}}}} = \frac{1}{{{z^{2023}}}}.\]

Do đó \[\frac{1}{{{x^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{{x^{2023}} + {y^{2023}} + {z^{2023}}}}.\]

⦁ Nếu \(y + z = 0\) hoặc \(x + z = 0,\) chứng minh tương tự ta cũng có

\[\frac{1}{{{x^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{{x^{2023}} + {y^{2023}} + {z^{2023}}}}.\]

Vậy nếu \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{{x + y + z}}\) thì \[\frac{1}{{{x^{2023}}}} + \frac{1}{{{y^{2023}}}} + \frac{1}{{{z^{2023}}}} = \frac{1}{{{x^{2023}} + {y^{2023}} + {z^{2023}}}}.\]