Question

Timf GTLN 

T= \(\dfrac{\text{8x+12}}{\text{x^2+4}}\)

Answer 1

\(T=\dfrac{8x+12}{x^2+4}=\dfrac{-\left(x^2+4\right)+\left(x^2+8x+16\right)}{x^2+4}\)

\(=\dfrac{\left(x+4\right)^2}{x^2+4}-1\text{≥}-1\)

Vậy Min\(=-1\text{⇔}x=-4\)

Answer 2

\(T=\dfrac{8x+12}{x^2+4}=\dfrac{4\left(x^2+4\right)-4\left(x^2-2x+1\right)}{x^2+4}\)

\(=-\dfrac{4\left(x-1\right)^2}{x^2+4}+4\text{≤}4\)

\(Max=4\)⇔\(x=1\)

Answer 3

Ta có: \(T=\dfrac{8x+12}{x^2+4}=\dfrac{4\left(x^2+4\right)-4\left(x^2-2x+1\right)}{x^2+4}=4-\dfrac{4\left(x-1\right)^2}{x^2+4}\le4\)

Dấu "=" xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)