Cho \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{a + b + c}}\). Chứng minh rằng:
\(\frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}} + {b^{2023}} + {c^{2023}}}}\).
Từ \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{a + b + c}}\), sử dụng quy tắc tính với phân thức, đa thức để rút gọn tìm ra a, b, c.
Theo đề bài ta có:
\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{a + b + c}}\)
\(\begin{array}{l}\frac{{bc + ac + ab}}{{abc}} = \frac{1}{{a + b + c}}\\\left( {bc + ac + ab} \right)\left( {a + b + c} \right) = abc\\bc\left( {a + b} \right) + b{c^2} + ac\left( {a + b} \right) + a{c^2} + ab\left( {a + b} \right) + abc - abc = 0\\bc\left( {a + b} \right) + ac\left( {a + b} \right) + ab\left( {a + b} \right) + \left( {b{c^2} + a{c^2}} \right) = 0\\bc\left( {a + b} \right) + ac\left( {a + b} \right) + ab\left( {a + b} \right) + {c^2}\left( {a + b} \right) = 0\\\left( {bc + ac + ab + {c^2}} \right)\left( {a + b} \right) = 0\\\left[ {\left( {bc + ab} \right) + \left( {ac + {c^2}} \right)} \right]\left( {a + b} \right) = 0\\\left[ {b\left( {a + c} \right) + c\left( {a + c} \right)} \right]\left( {a + b} \right) = 0\\\left( {b + c} \right)\left( {a + c} \right)\left( {a + b} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}b + c = 0\\a + c = 0\\a + b = 0\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}b = - c\\a = - c\\a = - b\end{array} \right.\end{array}\)
Trường hợp 1. Với \(b = - c\), ta có:
\(\begin{array}{l}VT = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}}\\ = \frac{1}{{{a^{2023}}}} + \frac{1}{{{{\left( { - c} \right)}^{2023}}}} + \frac{1}{{{c^{2023}}}}\\ = \frac{1}{{{a^{2023}}}} - \frac{1}{{{c^{2023}}}} + \frac{1}{{{c^{2023}}}}\\ = \frac{1}{{{a^{2023}}}}\end{array}\)
\(\begin{array}{l}VP = \frac{1}{{{a^{2023}} + {b^{2023}} + {c^{2023}}}}\\ = \frac{1}{{{a^{2023}} + {{\left( { - c} \right)}^{2023}} + {c^{2023}}}}\\ = \frac{1}{{{a^{2023}} - {c^{2023}} + {c^{2023}}}}\\ = \frac{1}{{{a^{2023}}}}\end{array}\)
\( \Rightarrow VT = VP\) hay \(\frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}} + {b^{2023}} + {c^{2023}}}}\)
Học sinh tự chứng minh tương tự cho trường hợp \(a = - c\) và \(a = - b\) .