Question

Đã trả lời bởi chuyên gia

Answer 1

$A = - x^{2} + 2 x + 7 = - (x^{2} - 2 x - 7) = - [(x^{2} - 2 x + 1) - 8] = - {(x - 1)}^{2} + 8 \leq 8 v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - 1 = 0 < = > x = 1 V ạ y v ớ i x = 1 t h ì A c ó G T L N = 8 B = 4 - x^{2} + 3 x = - (x^{2} - 3 x - 4) = - [(x^{2} - 3 x + \frac{9}{4}) - \frac{25}{4}] = - {(x - \frac{3}{2})}^{2} + \frac{25}{4} \leq \frac{25}{4} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - \frac{3}{2} = 0 < = > x = \frac{3}{2} V ậ y v ớ i x = \frac{3}{2} t h ì B c ó G T L N = \frac{25}{4}$

$C = - 4 x^{2} + x - 1 = - (4 x^{2} - x + 1) = - [(4 x^{2} - x + \frac{1}{16}) + \frac{15}{16}] = - {(2 x - \frac{1}{4})}^{2} - \frac{15}{16} \leq \frac{- 15}{16} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > 2 x - \frac{1}{4} = 0 < = > x = \frac{1}{8} V ậ y v ớ i x = \frac{1}{8} t h ì C c ó G T L N = \frac{- 15}{16} D = 5 x - 3 x^{2} + 6 = - 3 . (x^{2} - \frac{5}{3} x - 2) = - 3 . [(x^{2} - \frac{5}{3} x + \frac{25}{36}) - \frac{97}{36}] = - 3 . {(x - \frac{5}{6})}^{2} + \frac{97}{12} \leq \frac{97}{12} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - \frac{5}{6} = 0 < = > x = \frac{5}{6} V ậ y x = \frac{5}{6} t h ì D c ó G T L N = \frac{97}{12}$ .

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Answer 2

$A = - x^{2} + 2 x + 7 = - (x^{2} - 2 x - 7) = - [(x^{2} - 2 x + 1) - 8] = - {(x - 1)}^{2} + 8 \leq 8 v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - 1 = 0 < = > x = 1 V ạ y v ớ i x = 1 t h ì A c ó G T L N = 8 B = 4 - x^{2} + 3 x = - (x^{2} - 3 x - 4) = - [(x^{2} - 3 x + \frac{9}{4}) - \frac{25}{4}] = - {(x - \frac{3}{2})}^{2} + \frac{25}{4} \leq \frac{25}{4} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - \frac{3}{2} = 0 < = > x = \frac{3}{2} V ậ y v ớ i x = \frac{3}{2} t h ì B c ó G T L N = \frac{25}{4}$

$C = - 4 x^{2} + x - 1 = - (4 x^{2} - x + 1) = - [(4 x^{2} - x + \frac{1}{16}) + \frac{15}{16}] = - {(2 x - \frac{1}{4})}^{2} - \frac{15}{16} \leq \frac{- 15}{16} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > 2 x - \frac{1}{4} = 0 < = > x = \frac{1}{8} V ậ y v ớ i x = \frac{1}{8} t h ì C c ó G T L N = \frac{- 15}{16} D = 5 x - 3 x^{2} + 6 = - 3 . (x^{2} - \frac{5}{3} x - 2) = - 3 . [(x^{2} - \frac{5}{3} x + \frac{25}{36}) - \frac{97}{36}] = - 3 . {(x - \frac{5}{6})}^{2} + \frac{97}{12} \leq \frac{97}{12} v ớ i m ọ i x \in R D ấ u = x ả y r a < = > x - \frac{5}{6} = 0 < = > x = \frac{5}{6} V ậ y x = \frac{5}{6} t h ì D c ó G T L N = \frac{97}{12}$ .

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