Lời giải:\[\overline {abc} = 11\left( {{a^2} + {b^2} + {c^2}} \right) \Rightarrow \overline {abc} \vdots 11 \Rightarrow 100a + 10b + c \vdots 11 \Rightarrow 99a + 11b + a + c - b \vdots 11 \Rightarrow a + c - b \vdots 11\]
Lại có:
\[17 \ge a + c - b \ge - 8 \Rightarrow a + c - b \in \left\{ {0;11} \right\}\]
\[\begin{array}{l}
\boxed{TH1}:a + c - b = 11 \Leftrightarrow b + 11 = a + c \\
90 \ge {a^2} + {c^2} + {b^2} \ge \frac{{{{\left( {a + c} \right)}^2}}}{2} + {b^2} = \frac{{{{\left( {b + 11} \right)}^2}}}{2} + {b^2} \Leftrightarrow 3{b^2} + 22b - 59 \le 0 \\
\Leftrightarrow - 10 < \frac{{ - 11 - \sqrt {298} }}{3} \le b \le \frac{{ - 11 + \sqrt {298} }}{3} < 3 \Rightarrow b \in \left\{ {0;1;2} \right\} \\
*b = 0 \Rightarrow a = 11 - c \\
pt:\overline {a0c} = 11\left( {{c^2} + {a^2}} \right) \Leftrightarrow 100a + c = 11{c^2} + 11{a^2} \Leftrightarrow 100\left( {11 - c} \right) + c = 11{c^2} + 11{\left( {11 - c} \right)^2} \\
\Leftrightarrow 22{c^2} - 143c + 231 = 0 \Leftrightarrow \left[ \begin{array}{l}
c = 3 \\
c = \frac{7}{2} \\
\end{array} \right. \Rightarrow c = 3 \Rightarrow a = 8 \Rightarrow \boxed{\overline {abc} = 803} \\
*b = 1 \Rightarrow a = 12 - c \\
pt:\overline {a1c} = 11\left( {{c^2} + {a^2}} \right) \Leftrightarrow 100a + 10 + c = 11{c^2} + 11{a^2} \Leftrightarrow 100\left( {12 - c} \right) + 10 + c = 11{c^2} + 11{\left( {12 - c} \right)^2} \\
\Leftrightarrow 22{c^2} - 165c + 374 = 0:VN{\rm{ }}c \in N \\
*b = 2 \Rightarrow a = 13 - c \\
pt:\overline {a2c} = 11\left( {{c^2} + {a^2}} \right) \Leftrightarrow 100a + 20 + c = 11{c^2} + 11{a^2} \Leftrightarrow 100\left( {13 - c} \right) + 20 + c = 11{c^2} + 11{\left( {13 - c} \right)^2} \\
\Leftrightarrow 22{c^2} - 187c + 539 = 0:VN{\rm{ }}c \in N \\
\boxed{TH2}:a + c - b = 0 \Leftrightarrow b = a + c \\
pt:\overline {abc} = 11\left( {{a^2} + {b^2} + {c^2}} \right) \Leftrightarrow 100a + c + 10\left( {a + c} \right) = 11\left( {{a^2} + {c^2}} \right) + 11{\left( {a + c} \right)^2} \\
\Leftrightarrow 2{a^2} + 2ac - 10a + 2{c^2} - c = 0 \Leftrightarrow 2{a^2} + 2a\left( {c - 5} \right) + 2{c^2} - c = 0{\rm{ }}\left( 1 \right) \\
\Delta {'_a} = {\left( {c - 5} \right)^2} - 2\left( {2{c^2} - c} \right) = - 3{c^2} - 8c + 25 \ge 0 \Leftrightarrow \frac{{ - 4 - \sqrt {91} }}{3} \le c \le \frac{{\sqrt {91} - 4}}{3} < 2 \Rightarrow c \in \left\{ {0;1} \right\} \\
*c = 0 \Rightarrow \left( 1 \right):2{a^2} - 10a = 0 \Leftrightarrow \left[ \begin{array}{l}
a = 0 \\
a = 5 \\
\end{array} \right. \Rightarrow a = 5 \Rightarrow b = 5 \Rightarrow \boxed{\overline {abc} = 550} \\
*c = 1 \Rightarrow \left( 1 \right):2{a^2} - 8a + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}
a = \frac{{4 - \sqrt {14} }}{2}\not \in N \\
a = \frac{{4 + \sqrt {14} }}{2}\not \in N \\
\end{array} \right. \\
\end{array}\]
Kết luận:\[\overline {abc} \in \left\{ {550;803} \right\}\]
Bài viết đã được chỉnh sửa nội dung bởi perfectstrong: 09-01-2012 - 19:11