Question

bài 1. tìm GTLN của bỉu thức sau

2. B= -2x^2+8x-15

3. C=-3x^2+2x-1

4. D=-3x^2+4x-1

Answer 1

1)

B = -2x² + 8x - 15

= -2.(x² - 4x + 4) + 8 - 15

= -2.(x - 2)² - 7 \(\le\) - 7 với \(\forall\) x

Dấu " =" xảy ra khi (x - 2)² = 0 => x = 2

Vậy B_max = - 7 <=> x = 2

2)

C = - 3x² + 2x - 1

= - 3(x² - \(2.\dfrac{1}{3}\).x + \(\dfrac{1}{9}\) ) + \(\dfrac{1}{3}\) - 1

= - 3.(x - \(\dfrac{1}{3}\) )² - \(\dfrac{2}{3}\) \(\le\) - \(\dfrac{2}{3}\) với \(\forall\) x

Dấu " =" xảy ra khi (x - \(\dfrac{1}{3}\) )² = 0 => x = \(\dfrac{1}{3}\)

Vậy C_max = - \(\dfrac{2}{3}\) <=> x = \(\dfrac{1}{3}\)

3)

D = - 3x² + 4x - 1

= - 3(x² - \(2.\dfrac{2}{3}\).x + \(\dfrac{4}{9}\) ) - \(\dfrac{4}{3}\) - 1

= - 3.(x - \(\dfrac{2}{3}\) )² - \(\dfrac{7}{3}\) \(\le\) - \(\dfrac{7}{3}\) với \(\forall\) x

Dấu " =" xảy ra khi (x - \(\dfrac{2}{3}\) )² = 0 => x = \(\dfrac{2}{3}\)

Vậy D_max = - \(\dfrac{7}{3}\) <=> x = \(\dfrac{2}{3}\)

Answer 2

\(2.B=-2x^2+8x-15=-2\left(x^2-4x\right)-15\)

=> \(B=-\left(x-2\right)^2+8-15=-\left(x-2\right)^2-7\Leftrightarrow B_{Max}=-7\Leftrightarrow x=2\)

\(3.C=-3x^2+2x-1=-\left(3x^2-2x\right)-1\)

=> \(C=-3\left(x^2-\dfrac{2}{3}x\right)-1=-3\left(x-\dfrac{1}{3}\right)^2+\dfrac{1}{3}-1\)

=> \(C=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{2}{3}\Rightarrow C_{Max}=-\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)

\(4.D=-3x^2+4x-1=-\left(3x^2-4x\right)-1\)

=> \(D=-3\left(x^2-\dfrac{4}{3}x\right)-1=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{4}{3}-1=-3\left(x-\dfrac{2}{3}\right)^2-\dfrac{1}{3}\)

=> \(D_{Max}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{2}{3}\)