Đáp án:
\(\dfrac{3}{{2\left( {2x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
M = \left[ {\dfrac{{1 + \dfrac{1}{3}\left( {4x - 4\sqrt x + 1} \right) + 1 + \dfrac{1}{3}\left( {4x + 4\sqrt x + 1} \right)}}{{\left( {1 + \dfrac{1}{3}\left( {4x + 4\sqrt x + 1} \right)} \right)\left( {1 + \dfrac{1}{3}\left( {4x - 4\sqrt x + 1} \right)} \right)}}} \right].\dfrac{1}{{x + 1}}\\
= \dfrac{{2 + \dfrac{4}{3}x - \dfrac{4}{3}\sqrt x + \dfrac{1}{3} + \dfrac{4}{3}x + \dfrac{4}{3}\sqrt x + \dfrac{1}{3}}}{{\left( {1 + \dfrac{1}{3}\left( {4x + 4\sqrt x + 1} \right)} \right)\left( {1 + \dfrac{1}{3}\left( {4x - 4\sqrt x + 1} \right)} \right)}}.\dfrac{1}{{x + 1}}\\
= \dfrac{{\dfrac{8}{3} + \dfrac{8}{3}x}}{{\left( {1 + \dfrac{4}{3}x + \dfrac{4}{3}\sqrt x + \dfrac{1}{3}} \right)\left( {1 + \dfrac{4}{3}x - \dfrac{4}{3}\sqrt x + \dfrac{1}{3}} \right)}}.\dfrac{1}{{x + 1}}\\
= \dfrac{8}{3}:\left[ {\left( {\dfrac{4}{3}x + \dfrac{4}{3}\sqrt x + \dfrac{4}{3}} \right)\left( {\dfrac{4}{3}x - \dfrac{4}{3}\sqrt x + \dfrac{4}{3}} \right)} \right]\\
= \dfrac{8}{3}:\left[ {\dfrac{4}{3}\left( {x + \sqrt x + 1} \right).\dfrac{4}{3}\left( {x - \sqrt x + 1} \right)} \right]\\
= \dfrac{8}{3}:\dfrac{{16}}{9}\left( {x + \sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\\
= \dfrac{8}{3}.\dfrac{9}{{16\left[ {{{\left( {x + 1} \right)}^2} - x} \right]}}\\
= \dfrac{3}{{2\left[ {{x^2} + 2x + 1 - {x^2}} \right]}} = \dfrac{3}{{2\left( {2x + 1} \right)}}
\end{array}\)
