tìm GTNN,GTLN A=4-6X-X^2 B=3X^2-6X+1 C=5X^2-2X-3

Đáp án:

Giải thích các bước giải:

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