Đáp án:
b) Min=4
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:k = 1\\
Pt \to {x^2} - 6x - 8 = 0\\
\to \left[ \begin{array}{l}
x = 3 + \sqrt {17} \\
x = 3 - \sqrt {17}
\end{array} \right.\\
b)Xét:\Delta ' \ge 0\\
\to {k^2} + 4k + 4 + 2k + 6 \ge 0\\
\to {k^2} + 6k + 10 \ge 0\\
\to {k^2} + 6k + 9 + 1 \ge 0\\
\to {\left( {k + 3} \right)^2} + 1 \ge 0\left( {ld} \right)\forall k\\
Có:A = {\left( {{x_1} - {x_2}} \right)^2}\\
= {x_1}^2 - 2{x_1}{x_2} + {x_2}^2\\
= {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}\\
= {\left( {2k + 4} \right)^2} - 4\left( { - 2k - 6} \right)\\
= 4{k^2} + 16k + 16 + 8k + 24\\
= 4{k^2} + 24k + 40\\
= 4\left( {{k^2} + 6k + 10} \right)\\
= 4\left( {{k^2} + 6k + 9 + 1} \right)\\
= 4{\left( {k + 3} \right)^2} + 4\\
Do:4{\left( {k + 3} \right)^2} \ge 0\forall k\\
\to 4{\left( {k + 3} \right)^2} + 4 \ge 4\\
\to Min = 4\\
\Leftrightarrow k + 3 = 0\\
\to k = - 3
\end{array}\)
