Toán Lớp 8: Cho các số thực a, b, c thỏa mãn: a + b + c = 6 và ab + bc + ca = 12. Tính A =a mũ 4+b mũ 4+c mũ 4
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Toán Lớp 8: Cho các số thực a, b, c thỏa mãn: a + b + c = 6 và ab + bc + ca = 12. Tính A =a mũ 4+b mũ 4+c mũ 4
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Comments ( 2 )
a + b + c = 6\\
\Leftrightarrow {\left( {a + b + c} \right)^2} = {6^2}\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 36\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2}} \right) + 2.\left( {ab + bc + ca} \right) = 36\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2}} \right) + 2.12 = 36\\
\Leftrightarrow {a^2} + {b^2} + {c^2} = 12\\
\Rightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca\\
\Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} – 2ab – 2bc – 2ca = 0\\
\Leftrightarrow \left( {{a^2} – 2ab + {b^2}} \right) + \left( {{b^2} – 2bc + {c^2}} \right) + \left( {{c^2} – 2ca + {a^2}} \right) = 0\\
\Leftrightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} = 0\\
{\left( {a – b} \right)^2} \ge 0,\,\,\,\forall a,b\\
{\left( {b – c} \right)^2} \ge 0,\,\,\,\,\forall b,c\\
{\left( {c – a} \right)^2} \ge 0,\,\,\,\,\forall c,a\\
\Rightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} \ge 0,\,\,\,\,\forall a,b,c\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {a – b} \right)^2} = 0\\
{\left( {b – c} \right)^2} = 0\\
{\left( {c – a} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a – b = 0\\
b – c = 0\\
c – a = 0
\end{array} \right. \Rightarrow a = b = c\\
a + b + c = 6 \Rightarrow a = b = c = 2\\
\Rightarrow A = {a^4} + {b^4} + {c^4} = {2^4} + {2^4} + {2^4} = 48
\end{array}\)