Xem lại đề đi!!!!

Ta có: A = \(\dfrac{27-12x}{x^2+9}\) = \(\dfrac{\left(4x^2+36\right)-\left(4x^2+12x+9\right)}{x^2+9}\)

= \(\dfrac{4\left(x^2+9\right)-\left(2x+3\right)^2}{x^2+9}\)

= \(4-\dfrac{\left(2x+3\right)^2}{x^2+9}\)

Vì \(\left(2x+3\right)^2\) \(\ge\) 0

\(x^2+9\) > 0

=> \(\dfrac{\left(2x+3\right)^2}{x^2+9}\) \(\ge\) 0

=> \(4-\dfrac{\left(2x+3\right)^2}{x^2+9}\) \(\le\) 4

Dấu bằng xảy ra <=> \(\left(2x+3\right)^2\) = 0

<=> 2x +3 = 0

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

Vậy GTLN của A = 4 khi x = \(\dfrac{-3}{2}\)

 Tìm x biết: 2 ( x - 3 ) +  5 ( x + 4 ) = 49

<=> 2x - 6 + 5x +20 = 49

<=> 2x+ 5x = 49 + 6  -20

<=> 7x = 35

<=> x = 35:7

<=> x = 5