Cho a; b; c thoả mãn: \({a^{2022}}\; + {\rm{ }}{b^{2022}}\; + {\rm{ }}{c^{2022}}\; = {a^{1011}}{b^{1011}}{\rm{ + }}{b^{1011}}{c^{1011}}{\rm{ + }}{c^{1011}}{a^{1011}}\)
Tính giá trị của biểu thức \( \Rightarrow 2\left( {{a^{2022}}\; + {\rm{ }}{b^{2022}}\; + {\rm{ }}{c^{2022}}\;} \right) = 2\left( {{a^{1011}}{b^{1011}}{\rm{ + }}{b^{1011}}{c^{1011}}{\rm{ + }}{c^{1011}}{a^{1011}}} \right)\)
Dựa vào hằng đẳng thức \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\) để chứng minh.
Ta có: \({a^{2022}}\; + {\rm{ }}{b^{2022}}\; + {\rm{ }}{c^{2022}}\; = {a^{1011}}{b^{1011}}{\rm{ + }}{b^{1011}}{c^{1011}}{\rm{ + }}{c^{1011}}{a^{1011}}\)
\( \Rightarrow 2\left( {{a^{2022}}\; + {\rm{ }}{b^{2022}}\; + {\rm{ }}{c^{2022}}\;} \right) = 2\left( {{a^{1011}}{b^{1011}}{\rm{ + }}{b^{1011}}{c^{1011}}{\rm{ + }}{c^{1011}}{a^{1011}}} \right)\)
\(\left( {{a^{2022}} - 2{a^{1011}}{b^{1011}} + {b^{2022}}} \right) + \left( {{b^{2022}} - 2{b^{1011}}{c^{1011}} + {c^{2022}}} \right) + \left( {{c^{2022}} - 2{c^{1011}}{a^{1011}} + {a^{2022}}} \right) = 0\)
\( \Rightarrow {\left( {{a^{1011}} - {b^{1011}}} \right)^2} + {\left( {{b^{1011}} - {c^{1011}}} \right)^2} + {\left( {{c^{1011}} - {a^{1011}}} \right)^2} = 0\)
Vì \({x^2} \ge 0\) với \(\forall x\) nên dấu “=” xảy ra khi và chỉ khi
\({a^{1011}} - {b^{1011}} = {b^{1011}} - {c^{1011}} = {c^{1011}} - {a^{1011}} = 0 \Leftrightarrow a = b = c\)
\( \Rightarrow \) \(A = {\left( {a-b} \right)^{2020}} + {\left( {b-c} \right)^{2021}} + {\left( {a - c} \right)^{2022}} = 0\)
