-x^2+6 -11,x-x^2, 4x-x^2+3,5-8x-x^2,4x-x^2+1 03/07/2021 Bởi Athena -x^2+6 -11,x-x^2, 4x-x^2+3,5-8x-x^2,4x-x^2+1
Đáp án: \(\begin{array}{l}a)Max = 20\\b)Max = \dfrac{1}{4}\\c)Max = 7\\d)Max = 21\\e)Max = 5\end{array}\) Giải thích các bước giải: \(\begin{array}{l}a) – {x^2} + 6x – 11\\ = – \left( {{x^2} – 6x + 11} \right)\\ = – \left( {{x^2} – 6x + 9 – 20} \right)\\ = – {\left( {x – 3} \right)^2} + 20\\Do:{\left( {x – 3} \right)^2} \ge 0\forall x\\ \to – {\left( {x – 3} \right)^2} \le 0\\ \to – {\left( {x – 3} \right)^2} + 20 \le 20\\ \to Max = 20\\ \Leftrightarrow x = 3\\b)x – {x^2} = – \left( {{x^2} – x} \right)\\ = – \left( {{x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{1}{4}} \right)\\ = – {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{1}{4}\\Do:{\left( {x – \dfrac{1}{2}} \right)^2} \ge 0\forall x\\ \to – {\left( {x – \dfrac{1}{2}} \right)^2} \le 0\\ \to – {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\ \to Max = \dfrac{1}{4}\\ \Leftrightarrow x = \dfrac{1}{2}\\c)4x – {x^2} + 3 = – \left( {{x^2} – 4x – 3} \right)\\ = – \left( {{x^2} – 4x + 4 – 7} \right)\\ = – {\left( {x – 2} \right)^2} + 7\\Do:{\left( {x – 2} \right)^2} \ge 0\forall x\\ \to – {\left( {x – 2} \right)^2} \le 0\\ \to – {\left( {x – 2} \right)^2} + 7 \le 7\\ \to Max = 7\\ \Leftrightarrow x = 2\\d)5 – 8x – {x^2} = – \left( {{x^2} + 8x – 5} \right)\\ = – \left( {{x^2} + 8x + 16 – 21} \right)\\ = – {\left( {x + 4} \right)^2} + 21\\Do:{\left( {x + 4} \right)^2} \ge 0\forall x\\ \to – {\left( {x + 4} \right)^2} \le 0\\ \to – {\left( {x + 4} \right)^2} + 21 \le 21\\ \to Max = 21\\ \Leftrightarrow x = – 4\\e)4x – {x^2} + 1 = – \left( {{x^2} – 4x – 1} \right)\\ = – \left( {{x^2} – 4x + 4 – 5} \right)\\ = – {\left( {x – 2} \right)^2} + 5\\Do:{\left( {x – 2} \right)^2} \ge 0\forall x\\ \to – {\left( {x – 2} \right)^2} \le 0\\ \to – {\left( {x – 2} \right)^2} + 5 \le 5\\ \to Max = 5\\ \Leftrightarrow x = 2\end{array}\) Bình luận
Đáp án:
\(\begin{array}{l}
a)Max = 20\\
b)Max = \dfrac{1}{4}\\
c)Max = 7\\
d)Max = 21\\
e)Max = 5
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a) – {x^2} + 6x – 11\\
= – \left( {{x^2} – 6x + 11} \right)\\
= – \left( {{x^2} – 6x + 9 – 20} \right)\\
= – {\left( {x – 3} \right)^2} + 20\\
Do:{\left( {x – 3} \right)^2} \ge 0\forall x\\
\to – {\left( {x – 3} \right)^2} \le 0\\
\to – {\left( {x – 3} \right)^2} + 20 \le 20\\
\to Max = 20\\
\Leftrightarrow x = 3\\
b)x – {x^2} = – \left( {{x^2} – x} \right)\\
= – \left( {{x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{1}{4}} \right)\\
= – {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{1}{4}\\
Do:{\left( {x – \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to – {\left( {x – \dfrac{1}{2}} \right)^2} \le 0\\
\to – {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\to Max = \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{1}{2}\\
c)4x – {x^2} + 3 = – \left( {{x^2} – 4x – 3} \right)\\
= – \left( {{x^2} – 4x + 4 – 7} \right)\\
= – {\left( {x – 2} \right)^2} + 7\\
Do:{\left( {x – 2} \right)^2} \ge 0\forall x\\
\to – {\left( {x – 2} \right)^2} \le 0\\
\to – {\left( {x – 2} \right)^2} + 7 \le 7\\
\to Max = 7\\
\Leftrightarrow x = 2\\
d)5 – 8x – {x^2} = – \left( {{x^2} + 8x – 5} \right)\\
= – \left( {{x^2} + 8x + 16 – 21} \right)\\
= – {\left( {x + 4} \right)^2} + 21\\
Do:{\left( {x + 4} \right)^2} \ge 0\forall x\\
\to – {\left( {x + 4} \right)^2} \le 0\\
\to – {\left( {x + 4} \right)^2} + 21 \le 21\\
\to Max = 21\\
\Leftrightarrow x = – 4\\
e)4x – {x^2} + 1 = – \left( {{x^2} – 4x – 1} \right)\\
= – \left( {{x^2} – 4x + 4 – 5} \right)\\
= – {\left( {x – 2} \right)^2} + 5\\
Do:{\left( {x – 2} \right)^2} \ge 0\forall x\\
\to – {\left( {x – 2} \right)^2} \le 0\\
\to – {\left( {x – 2} \right)^2} + 5 \le 5\\
\to Max = 5\\
\Leftrightarrow x = 2
\end{array}\)