Question

Tìm giá trị lớn nhất của biểu thức P = \(-2x^2+6x+9\)

Answer 1

`P=-2x^2 +6x+9`

`=-2(x^2 -3x-9/2)`

`=-2[(x^2 - 2. 3/2 .x + 9/4) - 27/4]`

`=-2[(x -3/2)^2 - 27/4]`

`=-2(x -3/2)^2 + 27/2 <= 27/2`

Dấu "=" xảy ra `<=>x -3/2 = 0 <=> x=3/2`

Vậy `P_(max)= 27/2 <=> x= 3/2`

Answer 2

\(P=-2x^2+6x+9\)

\(P=-\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{27}{2}\)

\(P=-\left[\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3\sqrt{2}}{2}+\left(\dfrac{3\sqrt{2}}{2}\right)^2\right]+\dfrac{27}{2}\)

\(P=-\left(\sqrt{2}x-\dfrac{3\sqrt{2}}{2}\right)^2+\dfrac{27}{2}\le\dfrac{27}{2}\)

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

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(Max_P=\dfrac{27}{2}\) khi \(x=\dfrac{3}{2}\)

 

 

Answer 3

`= -2(x^2 - 3x - 4,5)`

`= -2(x^2 - 2 . 3/2x + 9/4 - 27/4 )`

`= -2((x-3/2)^2 - 27/4)`

Vì `(x-3/2)^2 >=0 => (x-3/2) - 27/4 >= 0 - 27/4 = -27/4`

`=> Mi``n P = -27/4 xx -2 = 27/2`.

Dấu bằng xảy ra `<=> x = 3/2`.