Đáp án:
\(\begin{array}{l}
1)Min = - 2\\
2)Min = 2\\
3)Max = - 5\\
4)Max = - 31\\
5)Min = \dfrac{3}{4}\\
6)Max = 21
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)9{x^2} + 6x - 1\\
= 9{x^2} + 6x + 1 - 2\\
= {\left( {3x + 1} \right)^2} - 2\\
Do:{\left( {3x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {3x + 1} \right)^2} - 2 \ge - 2\\
\to Min = - 2\\
\Leftrightarrow x = - \dfrac{1}{3}\\
2)4{x^2} - 4x + 1 + 2\\
= {\left( {2x + 1} \right)^2} + 2\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 1} \right)^2} + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow x = - \dfrac{1}{2}\\
3) - {x^2} + 10x - 30\\
= - \left( {{x^2} - 10x + 25 + 5} \right)\\
= - {\left( {x - 5} \right)^2} - 5\\
Do:{\left( {x - 5} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 5} \right)^2} \le 0\\
\to - {\left( {x - 5} \right)^2} - 5 \le - 5\\
\to Max = - 5\\
4)25{x^2} + 10x - 30\\
= 25{x^2} + 10x + 1 - 31\\
= - {\left( {5x + 1} \right)^2} - 31\\
Do:{\left( {5x + 1} \right)^2} \ge 0\forall x\\
\to - {\left( {5x + 1} \right)^2} \le 0\\
\to - {\left( {5x + 1} \right)^2} - 31 \le - 31\\
\to Max = - 31\\
\Leftrightarrow x = - \dfrac{1}{5}\\
5){x^2} - x + 1 = {x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to Min = \dfrac{3}{4}\\
\to x = \dfrac{1}{2}\\
6) - {x^2} + 8x + 5\\
= - \left( {{x^2} - 8x - 5} \right)\\
= - \left( {{x^2} - 8x + 16 - 21} \right)\\
= - {\left( {x - 4} \right)^2} + 21\\
Do:{\left( {x - 4} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 4} \right)^2} \le 0\\
\to - {\left( {x - 4} \right)^2} + 21 \le 21\\
\to Max = 21\\
\Leftrightarrow x = 4
\end{array}\)
