Đáp án:
\(\left[ \begin{array}{l}
a = \frac{{2\sqrt {15} }}{3}\\
a = - \frac{{2\sqrt {15} }}{3}
\end{array} \right. \to \left[ \begin{array}{l}
b = - \frac{7}{3}\\
b = - \frac{7}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x_1} - {x_2} = 4\\
\left( {{x_1} - {x_2}} \right)\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right) = 28
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} - {x_2} = 4\\
4\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right) = 28\left( * \right)
\end{array} \right.\\
\left( * \right) \to {x_1}^2 + {x_1}{x_2} + {x_2}^2 = 7\\
\to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + {x_1}{x_2} = 7\\
\to {\left( {{x_1} + {x_2}} \right)^2} - {x_1}{x_2} = 7\\
\to {a^2} - b - 2 = 7\\
\to b = {a^2} - 9\\
Thay:b = {a^2} - 9\\
Pt \to {x^2} + ax + {a^2} - 9 = 0\left( 1 \right)
\end{array}\)
Để phương trình (1) có 2 nghiệm phân biệt
⇔ Δ>0
\(\begin{array}{l}
\to {a^2} - 4{a^2} + 36 > 0\\
\to 36 - 3{a^2} > 0\\
\to a \in \left( { - 2\sqrt 3 ;2\sqrt 3 } \right)\\
\to \left[ \begin{array}{l}
x = \frac{{ - a + \sqrt {36 - 3{a^2}} }}{2}\\
x = \frac{{ - a - \sqrt {36 - 3{a^2}} }}{2}
\end{array} \right.\\
Có:{x_1} - {x_2} = 4\\
\to \frac{{ - a + \sqrt {36 - 3{a^2}} }}{2} - \frac{{ - a - \sqrt {36 - 3{a^2}} }}{2} = 4\\
\to - a + \sqrt {36 - 3{a^2}} + a + \sqrt {36 - 3{a^2}} = 8\\
\to 2\sqrt {36 - 3{a^2}} = 8\\
\to 36 - 3{a^2} = 16\\
\to 3{a^2} - 20 = 0\\
\to \left[ \begin{array}{l}
a = \frac{{2\sqrt {15} }}{3}\\
a = - \frac{{2\sqrt {15} }}{3}
\end{array} \right.\left( {TM} \right) \to \left[ \begin{array}{l}
b = - \frac{7}{3}\\
b = - \frac{7}{3}
\end{array} \right.
\end{array}\)
