Question

[Lâm Ánh Yên]

Tìm GTLN của:

\(2x-2x^2-5\)

Answer 1

[Dũng Nguyễn]

Ta có:\(2x-2x^2-5=-\left(2x^2-2x+5\right)\)

\(=-\left[2\left(x^2-x+\dfrac{5}{2}\right)\right]\)

\(=-\left\{2\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+\dfrac{5}{2}\right]\right\}\)

\(=-\left\{2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\right\}\)

\(=-\left[2\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{2}\right]\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)

Do \(-2\left(x-\dfrac{1}{2}\right)^2\le0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\) )

\(\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\) hay \(2x-2x^2-5\le-\dfrac{9}{2}\) (dấu ''='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

Vậy giá trị lớn nhất của biểu thức \(2x-2x^2-5\) là \(-\dfrac{9}{2}\) tại \(x=\dfrac{1}{2}\)

Answer 2

[Lê Minh Anh]

A = 2x - 2x² - 5

=> 2A = -4x² + 4x - 10

=> 2A = -(4x² - 4x + 10)

=> 2A = - [(2x)² - 2.2x + 1] - 9

=> 2A = -(2x - 1)² -9

Mà: -(2x - 1)² \(\le\) 0 => -(2x - 1)² - 9 \(\le\) -9

=> 2A \(\le\) -9

=> A \(\le\) -4,5

Đẳng thức xảy ra khi: -(2x - 1)² = 0 <=> x = \(\dfrac{1}{2}\)