$\dfrac{12x^2+12x+11}{4x^2+4x+3}=\dfrac{5y^2-10y+9}{y^2-2y+2}$
Xét $\dfrac{12x^2+12x+11}{4x^2+4x+3}=\dfrac{12x^2+12x+9}{4x^2+4x+3}+\dfrac{2}{4x^2+4x+3}$
$\iff \dfrac{12x^2+12x+11}{4x^2+4x+3}=3+\dfrac{2}{(2x+1)^2+2}$
$\iff \dfrac{12x^2+12x+11}{4x^2+4x+3} \le 3+\dfrac{2}{2}$
$\iff \dfrac{12x^2+12x+11}{4x^2+4x+3} \le 4\ \bigstar$
Xét $\dfrac{5y^2-10y+9}{y^2-2y+2}=\dfrac{5y^2-10y+10}{y^2-2y+2}-\dfrac{1}{y^2-2y+2}$
$\iff \dfrac{5y^2-10y+9}{y^2-2y+2}=5-\dfrac{1}{(y-1)^2+1}$
$\iff \dfrac{5y^2-10y+9}{y^2-2y+2} \ge 5-\dfrac{1}{1}$
$\iff \dfrac{5y^2-10y+9}{y^2-2y+2} \ge 4\ \bigstar \bigstar$
Từ $\bigstar$ và $\bigstar \bigstar$, ta có: $\dfrac{12x^2+12x+11}{4x^2+4x+3} \le
\dfrac{5y^2-10y+9}{y^2-2y+2}$
Dấu "=" xảy ra khi $x=\dfrac{-1}{2},\ y=1$
Vậy ....................................