\(\dfrac{5}{x+2}\) ≤ 5

Max A=5 ⇔x+2=1

               ⇔x=-1

Biểu thức nào em?

`M = (x^2 + 5 - 2x + 4)/(x^2+5)`

`= 1 - (2x-4)/(x^2+5) <= 1 - 0 = 1 (x^2+5 >0)`.

Dấu `=` xảy ra `<=> 2x- 4 = 0 <=> x = 2.`

Vậy ...

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)