Đáp án:

\({{27} \over {64}}\)

Giải thích các bước giải:

$$\eqalign{ & {\left( {{1 \over 4} + {3 \over 4}x} \right)^4} = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 - k}}{{\left( {{3 \over 4}x} \right)}^k}} \cr & = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 - k}}{{\left( {{3 \over 4}} \right)}^k}{x^k}} \cr & Gia\,\,su\,\,{T_{k + 1}} = C_4^k{\left( {{1 \over 4}} \right)^{4 - k}}{\left( {{3 \over 4}} \right)^k}\,\,la\,\,he\,\,so\,\,lon\,\,nhat \cr & \Leftrightarrow \left\{ \matrix{ C_4^k{\left( {{1 \over 4}} \right)^{4 - k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k + 1}{\left( {{1 \over 4}} \right)^{3 - k}}{\left( {{3 \over 4}} \right)^{k + 1}} \hfill \cr C_4^k{\left( {{1 \over 4}} \right)^{4 - k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k - 1}{\left( {{1 \over 4}} \right)^{5 - k}}{\left( {{3 \over 4}} \right)^{k - 1}} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {{4!} \over {k!\left( {4 - k} \right)!}}{1 \over 4} > {{4!} \over {\left( {k + 1} \right)!\left( {3 - k} \right)!}}{3 \over 4} \hfill \cr {{4!} \over {k!\left( {4 - k} \right)!}}{3 \over 4} > {{4!} \over {\left( {k - 1} \right)!\left( {5 - k} \right)!}}{1 \over 4} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {1 \over {4 - k}} > {3 \over {k + 1}} \hfill \cr {3 \over k} > {1 \over {5 - k}} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ k + 1 > 12 - 3k \hfill \cr 15 - 3k > k \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 4k > 11 \hfill \cr 4k < 15 \hfill \cr} \right. \cr & \Leftrightarrow {{11} \over 4} < k < {{15} \over 4}.\,\,\,Ma\,\,k \in Z \Rightarrow k = 3\,\,\left( {tm} \right) \cr & Vay\,\,he\,\,so\,\,lon\,\,nhat\,\,la\,\,C_4^3\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}\, = {{27} \over {64}}\, \cr} $$