$\begin{array}{l}\quad A = \dfrac{y^4 +12y^2 + 11}{y^4 + 6y^2 + 5}\\ a)\quad A = \dfrac{y^4 + y^2 + 11y^2 + 11}{y^4 + y^2 + 5y^2 + 5}\\ \to A = \dfrac{y^2(y^2 + 1) + 11(y^2 + 1)}{y^2(y^2 +1) + 5(y^2 + 1)}\\ \to A = \dfrac{(y^2+1)(y^2 +11)}{(y^2+1)(y^2+5)}\\ \to A = \dfrac{y^2+11}{y^2+5}\\ b)\quad A = \dfrac{y^2 +11}{y^2 + 5}\\ \to A = \dfrac{y^2 + 5 + 6}{y^2 + 5}\\ \to A = 1 + \dfrac{6}{y^2 + 5}\\ \text{Ta có:}\\ \quad y^2 \leq 0 \quad \forall y\\ \to y^2 + 5 \geq 5\\ \to \dfrac{6}{y^2+ 5} \leq \dfrac{6}{5}\\ \to 1 + \dfrac{6}{y^2 + 5} \leq \dfrac{11}{5}\\ \to A \leq \dfrac{11}{5}\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow y^2 = 0 \Leftrightarrow y = 0\\ Vậy\,\,\max A = \dfrac{11}{5} \Leftrightarrow y = 0 \end{array}$
