Giải thích các bước giải:
\(\begin{array}{l}
1)a)A = \dfrac{{\left( {6x + 1} \right)\left( {x + 6} \right) + \left( {6x - 1} \right)\left( {x - 6} \right)}}{{x\left( {x - 6} \right)\left( {x + 6} \right)}}.\dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{12\left( {{x^2} + 1} \right)}}\\
= \dfrac{{6{x^2} + 7x + 6 + 6{x^2} - 7x + 6}}{{12x\left( {{x^2} + 1} \right)}}\\
= \dfrac{{12{x^2} + 12}}{{x\left( {12{x^2} + 12} \right)}} = \dfrac{1}{x}\\
2b){x^2} + 3{y^2} = 4xy\\
\Leftrightarrow {x^2} - 4xy + 3{y^2} = 0\\
\Leftrightarrow {x^2} - xy - 3xy + 3{y^2} = 0\\
\Leftrightarrow x\left( {x - y} \right) - 3y\left( {x - y} \right) = 0\\
\Leftrightarrow \left( {x - 3y} \right)\left( {x - y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3y\\
x = y\left( L \right)
\end{array} \right.\\
\Rightarrow A = \dfrac{{2.3y + 5y}}{{3y - 2y}} = \dfrac{{11y}}{y} = 11\\
3b)M = 1 - \dfrac{4}{x} + \dfrac{{2013}}{{{x^2}}}\\
= 2013\left( {\dfrac{1}{{{x^2}}} - 2.\dfrac{2}{{2013}}x + \dfrac{4}{{{{2013}^2}}}} \right) + \dfrac{{2009}}{{2013}}\\
= 2013{\left( {\dfrac{1}{x} - \dfrac{2}{{2013}}} \right)^2} + \dfrac{{2009}}{{2013}} \ge \dfrac{{2009}}{{2013}}\\
{M_{\min }} = \dfrac{{2009}}{{2013}} \Leftrightarrow \dfrac{1}{x} = \dfrac{2}{{2013}} \Leftrightarrow x = \dfrac{{2013}}{2}
\end{array}\)
Em tách ra hỏi từng câu thôi nhé
