$E = \dfrac{x^2-x+10}{x-3} $
$=\dfrac{(x-3)^2 + 5.(x-3) +16}{x-3} $
$=(x-3)+\dfrac{16}{x-3} + 5 ≥2\sqrt[]{16} + 5 = 2.4+5 = 13$
Dấu "=" xảy ra $⇔x-3=4 ⇔x=7$
$F = \dfrac{x^2+7}{x+3} $
$=\dfrac{(x+3)^2-6.(x+3)+16}{x+3} $
$=(x+3)+\dfrac{16}{x+3} - 6 ≥ 2\sqrt[]{16} - 6 = 2 $
Dấu "=" xảy ra $⇔x+3=4 ⇔x=1$
$G = \dfrac{x^2}{x-1}$
$=\dfrac{x^2-1+1}{x-1} $
$=x+1+\dfrac{1}{x-1} $
$=(x-1+\dfrac{1}{x-1}) + 2 ≥2\sqrt[]{1}+2 = 4$
Dấu "=" xảy ra $⇔x-1=1⇔x=2$
