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