tìm gtln,gtnn a=4-6x-x^2 b=3x^2-6x+1 c=5x^2-2x-3 02/10/2021 Bởi Natalia tìm gtln,gtnn a=4-6x-x^2 b=3x^2-6x+1 c=5x^2-2x-3
Đáp án: Giải thích các bước giải: $\begin{array}{l} a = 4 – 6x – {x^2}\\ = 13 – 9 – 6x – {x^2} = 13 – \left( {9 + 6x + {x^2}} \right)\\ = 13 – {\left( {3 + x} \right)^2}\\ Vi\,\,{\left( {3 + x} \right)^2} \ge 0\, \Rightarrow 13 – {\left( {x + 3} \right)^2} \le 13\\ GTLN\,cua\,a\,la\,\,13 \Leftrightarrow x = – 3\\ b = 3{x^2} – 6x + 1 = 3{x^2} – 6x + 3 – 2\\ = 3\left( {{x^2} – 2x + 1} \right) – 2 = 3{\left( {x – 1} \right)^2} – 1\\ Vi\,\,{\left( {x – 1} \right)^2} \ge 0\, \Rightarrow 3{\left( {x – 1} \right)^2} – 2 \ge – 2\\ GTNN\,cua\,\,b\,la\,\, – 2\,\,khi\,\,x = 1\\ c = 5{x^2} – 2x – 3 = {\left( {\sqrt 5 x} \right)^2} – 2.\sqrt 5 x.\dfrac{1}{{\sqrt 5 }} + \dfrac{1}{5} – \dfrac{{16}}{5}\\ = {\left( {\sqrt 5 x – \dfrac{1}{{\sqrt 5 }}} \right)^2} – \dfrac{{16}}{5} \ge – \dfrac{{16}}{5}\\ GTNN\,cua\,c\,la\,\, – \dfrac{{16}}{5} \Leftrightarrow x = \dfrac{1}{5} \end{array}$ Bình luận
Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
a = 4 – 6x – {x^2}\\
= 13 – 9 – 6x – {x^2} = 13 – \left( {9 + 6x + {x^2}} \right)\\
= 13 – {\left( {3 + x} \right)^2}\\
Vi\,\,{\left( {3 + x} \right)^2} \ge 0\, \Rightarrow 13 – {\left( {x + 3} \right)^2} \le 13\\
GTLN\,cua\,a\,la\,\,13 \Leftrightarrow x = – 3\\
b = 3{x^2} – 6x + 1 = 3{x^2} – 6x + 3 – 2\\
= 3\left( {{x^2} – 2x + 1} \right) – 2 = 3{\left( {x – 1} \right)^2} – 1\\
Vi\,\,{\left( {x – 1} \right)^2} \ge 0\, \Rightarrow 3{\left( {x – 1} \right)^2} – 2 \ge – 2\\
GTNN\,cua\,\,b\,la\,\, – 2\,\,khi\,\,x = 1\\
c = 5{x^2} – 2x – 3 = {\left( {\sqrt 5 x} \right)^2} – 2.\sqrt 5 x.\dfrac{1}{{\sqrt 5 }} + \dfrac{1}{5} – \dfrac{{16}}{5}\\
= {\left( {\sqrt 5 x – \dfrac{1}{{\sqrt 5 }}} \right)^2} – \dfrac{{16}}{5} \ge – \dfrac{{16}}{5}\\
GTNN\,cua\,c\,la\,\, – \dfrac{{16}}{5} \Leftrightarrow x = \dfrac{1}{5}
\end{array}$