Đề ở đâu vậy bạn?
Đây là lời giải của mình cho bài viete bậc 3
còn bài kia thì cũng bình thường.
Theo định lý Viete, ta có:
\[\left\{ \begin{array}{l}
{x_1} + {x_2} + {x_3} = 17 \\
{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = a \\
{x_1}{x_2}{x_3} = {b^2} \\
\end{array} \right. \Rightarrow {x_1};{x_2};{x_3} > 0 \Rightarrow {x_1} \ge {x_2} \ge {x_3} \ge 1\]
\[\begin{array}{l}
a \le \frac{{{{\left( {{x_1} + {x_2} + {x_3}} \right)}^2}}}{3} < 97 \Rightarrow 1 \le a \le 96 \\
{x_1} + {x_2} + {x_3} \ge 3\sqrt[3]{{{x_1}{x_2}{x_3}}} \Rightarrow {b^2} \le {\left( {\frac{{17}}{3}} \right)^3} < 196 \Rightarrow |b| < 14 \Rightarrow 1 \le b \le 13 \\
{x_1}|b;{x_1} \le 15;x_1^3 \ge {b^2} \Rightarrow {x_1} \ge \sqrt[3]{{{b^2}}} \\
\end{array}\]
Ta xét các TH:
\[\begin{array}{l}
TH1:\left( {{x_1};{x_2};{x_3}} \right) = \left( {{b^2};1;1} \right) \Rightarrow {b^2} = 15:False \\
TH2:\left( {{x_1};{x_2};{x_3}} \right) = \left( {b;b;1} \right) \Rightarrow b = 8 \Rightarrow a = 80 \\
TH3:Other \\
*b = 1 \Rightarrow {x_1} = {x_2} = {x_3} = 1:False \\
*b \in \left\{ {2;3;5;7;11;13} \right\} \Rightarrow \left( {{x_1};{x_2};{x_3}} \right) = \left( {{b^2};1;1} \right);\left( {b;b;1} \right):False \\
*b = 4 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{4^2}}} > 2 \Rightarrow \left( {{x_1};{x_2};{x_3}} \right) = \left( {8;2;1} \right);\left( {4;4;1} \right):False \\
*b = 8 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{8^2}}} = 4 \Rightarrow \left( {{x_1};{x_2};{x_3}} \right) = \left( {8;4;2} \right);\left( {4;4;4} \right):False \\
*b = 6 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{6^2}}} > 3 \Rightarrow \left( {{x_1};{x_2};{x_3}} \right) = \left( {12;3;1} \right);\left( {9;4;1} \right);\left( {9;2;2} \right);\left( {6;3;2} \right);\left( {4;3;3} \right):False \\
*b = 9 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{9^2}}} > 4 \Rightarrow \left( {{x_1};{x_2};{x_3}} \right) = \left( {9;3;3} \right):False \\
*b = 10 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{{10}^2}}} > 4\left[ \begin{array}{l}
\left( {{x_1};{x_2};{x_3}} \right) = \left( {10;5;2} \right):True \Rightarrow a = 80 \\
\left( {{x_1};{x_2};{x_3}} \right) = \left( {5;5;4} \right):False \\
\end{array} \right. \\
*b = 12 \Rightarrow 15 \ge {x_1} \ge \sqrt[3]{{{{12}^2}}} > 5 \Rightarrow \left[ \begin{array}{l}
\left( {{x_1};{x_2};{x_3}} \right) = \left( {12;6;2} \right);\left( {12;4;3} \right);\left( {9;8;2} \right):False \\
\left( {{x_1};{x_2};{x_3}} \right) = \left( {9;4;4} \right):True \Rightarrow a = 88 \\
\end{array} \right. \\
\left( {a;b} \right) = \left( {80;8} \right);\left( {80;10} \right);\left( {88;12} \right) \\
\end{array}\]
