Đáp án:
\[{a^4} + {b^4} + {c^4} = 2020050\]
Giải thích các bước giải:
\({\left( {x + y + z} \right)^2} = \left( {{x^2} + {y^2} + {z^2}} \right) + 2.\left( {xy + yz + zx} \right)\)
Ta có:
\(\begin{array}{l}
a + b + c = 0\\
\Leftrightarrow {\left( {a + b + c} \right)^2} = 0\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2}} \right) + 2.\left( {ab + bc + ca} \right) = 0\\
\Leftrightarrow 2010 + 2.\left( {ab + bc + ca} \right) = 0\\
\Leftrightarrow ab + bc + ca = - 1005\\
\Leftrightarrow {\left( {ab + bc + ca} \right)^2} = {1005^2}\\
\Leftrightarrow \left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) + 2.\left( {ab.bc + bc.ca + ca.ab} \right) = {1005^2}\\
\Leftrightarrow \left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) + 2.abc.\left( {a + b + c} \right) = {1005^2}\\
\Leftrightarrow {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} = {1005^2}\,\,\,\,\,\,\,\left( {a + b + c = 0} \right)\\
{a^2} + {b^2} + {c^2} = 2010\\
\Leftrightarrow {\left( {{a^2} + {b^2} + {c^2}} \right)^2} = {2010^2}\\
\Leftrightarrow \left( {{a^4} + {b^4} + {c^4}} \right) + 2.\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = {2010^2}\\
\Leftrightarrow \left( {{a^4} + {b^4} + {c^4}} \right) + {2.1005^2} = {2010^2}\\
\Leftrightarrow {a^4} + {b^4} + {c^4} = 2020050
\end{array}\)
Vậy \({a^4} + {b^4} + {c^4} = 2020050\)
