Toán Lớp 8: Cho các số thực a,b,c thỏa mãn a+b+c=ab+bc+ca=3 Chứng minh rằng a=b=c=1
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Toán Lớp 8: Cho các số thực a,b,c thỏa mãn a+b+c=ab+bc+ca=3 Chứng minh rằng a=b=c=1
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a + b + c = 3\\
\Leftrightarrow {\left( {a + b + c} \right)^2} = {3^2}\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 9\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2}} \right) + 2.\left( {ab + bc + ca} \right) = 9\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2}} \right) + 2.3 = 9\\
\Leftrightarrow {a^2} + {b^2} + {c^2} = 3\\
\Rightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca\\
\Rightarrow 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. \begin{array}{l}
{\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
\end{array} \right\} \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\\
b = c\\
c = a
\end{array} \right. \Rightarrow a = b = c = \frac{{a + b + c}}{3} = 1\\
\Rightarrow a = b = c = 1
\end{array}\)