Question

1) Tìm giá trị nhỏ nhất của :

D = 3² - 7x + 6

E = 4x² - 7x + 8

G = 5x² + 3x + 5

2) Tìm giá trị lớn nhất của :

A = -4x² - 3x + 7

B = -5x² + 8x - 3

C = -7x² + 6x + 5

D = -x² - 3x + 6

Answer 1

1) D = x² - 7x + 6

D = x² - 2.\(\dfrac{7}{2}x+\dfrac{49}{4}+6-\dfrac{49}{4}\)

D = \(\left(x-\dfrac{7}{2}\right)^2\) - \(\dfrac{25}{4}\)

Do : \(\left(x-\dfrac{7}{2}\right)^2\)≥ 0 ∀x

⇒ \(\left(x-\dfrac{7}{2}\right)^2\) - \(\dfrac{25}{4}\) ≥ - \(\dfrac{25}{4}\)

⇒D_MIN = - \(\dfrac{25}{4}\) ⇔ x = \(\dfrac{7}{2}\)

E = 4x² - 7x + 8

E = 4( x² - \(2.\dfrac{7}{8}x+\dfrac{49}{64}\)) - \(\dfrac{49}{16}\) + 8

E = 4( x - \(\dfrac{7}{8}\))² + \(\dfrac{79}{16}\) ≥ \(\dfrac{79}{16}\)

⇒ E_MIN = \(\dfrac{79}{16}\) ⇔ x = \(\dfrac{7}{8}\)

Answer 2

Làm mẫu:

Câu 1:

\(G=5x^2-7x+8\\ =5x^2-7x+\dfrac{49}{20}+\dfrac{111}{20}\\ =\left(5x^2-7x+\dfrac{49}{20}\right)+\dfrac{111}{20}\\ =5\left(x^2-\dfrac{7}{5}x+\dfrac{49}{100}\right)+\dfrac{111}{20}\\ =5\left(x-\dfrac{7}{10}\right)^2+\dfrac{111}{20}\)

Do \(\left(x-\dfrac{7}{10}\right)^2\ge0\forall x\)

\(\Rightarrow5\left(x-\dfrac{7}{10}\right)^2\ge0\forall x\\ \Rightarrow G=5\left(x-\dfrac{7}{10}\right)^2+\dfrac{111}{20}\ge\dfrac{111}{20}\forall x\)

Dấu \("="\) xảy ra khi:

\(\left(x-\dfrac{7}{10}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{10}=0\\ \Leftrightarrow x=\dfrac{7}{10}\)

Vậy \(G_{Min}=\dfrac{111}{20}\) khi \(x=\dfrac{7}{10}\)

Câu 2:

\(C=-7x^2+6x+5\\ =-7x^2+6x-\dfrac{9}{7}+\dfrac{44}{7}\\ =-\left(7x^2-6x+\dfrac{9}{7}\right)+\dfrac{44}{7}\\ =-7\left(x^2-\dfrac{6}{7}x+\dfrac{9}{49}\right)+\dfrac{44}{7}\\ =-7\left(x-\dfrac{3}{7}\right)^2+\dfrac{44}{7}\)

Do \(-7\left(x-\dfrac{3}{7}\right)^2\le0\forall x\)

\(\Rightarrow C=-7\left(x-\dfrac{3}{7}\right)^2+\dfrac{44}{7}\le\dfrac{44}{7}\forall x\)

Dấu \("="\) xảy ra khi:

\(-7\left(x-\dfrac{3}{7}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{7}=0\\ \Leftrightarrow x=\dfrac{3}{7}\)

Vậy \(C_{Max}=\dfrac{44}{7}\) khi \(x=\dfrac{3}{7}\)