Cho a^2022+b^2022+c^2022=a^1011b^1011+b^1011c^1011+c^1011a^1011 Tính M = (a-b)^2020+(b-c)^2021+(c-a)^2022

Đáp án: $M = 0$

Giải thích các bước giải:

$\begin{array}{l}
{a^{2022}} + {b^{2022}} + {c^{2022}}\\
 = {a^{1011}}.{b^{1011}} + {b^{1011}}.{c^{1011}} + {c^{1011}}.{a^{1011}}\\
 \Leftrightarrow 2{a^{2022}} + 2{b^{2022}} + 2{c^{2022}}\\
 = 2{a^{1011}}.{b^{1011}} + 2{b^{1011}}.{c^{1011}} + 2{c^{1011}}.{a^{1011}}\\
 \Leftrightarrow {a^{2022}} - 2{a^{1011}}.{b^{1011}} + {b^{2022}}\\
 + {b^{2022}} - 2{b^{1011}}.{c^{1011}} + {c^{2022}}\\
 + {c^{2022}} - 2{c^{1011}}.{a^{1011}} + {a^{2022}} = 0\\
 \Leftrightarrow {\left( {{a^{1011}} - {b^{1011}}} \right)^2} + {\left( {{b^{1011}} - {c^{1011}}} \right)^2} + {\left( {{c^{1011}} - {a^{1011}}} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {{a^{1011}} - {b^{1011}}} \right)^2} \ge 0\\
{\left( {{b^{1011}} - {c^{1011}}} \right)^2} \ge 0\\
{\left( {{c^{1011}} - {a^{1011}}} \right)^2} \ge 0
\end{array} \right.\\
 \Leftrightarrow {\left( {{a^{1011}} - {b^{1011}}} \right)^2} + {\left( {{b^{1011}} - {c^{1011}}} \right)^2} + {\left( {{c^{1011}} - {a^{1011}}} \right)^2} \ge 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {{a^{1011}} - {b^{1011}}} \right)^2} = 0\\
{\left( {{b^{1011}} - {c^{1011}}} \right)^2} = 0\\
{\left( {{c^{1011}} - {a^{1011}}} \right)^2} = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{a^{1011}} = {b^{1011}}\\
{b^{1011}} = {c^{1011}}\\
{c^{1011}} = {a^{1011}}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = b\\
b = c\\
c = a
\end{array} \right.\\
 \Leftrightarrow a = b = c\\
M = {\left( {a - b} \right)^{2020}} + {\left( {b - c} \right)^{2021}} + {\left( {c - a} \right)^{2022}} = 0
\end{array}$