Toán Hệ số lớn nhất trong khai triển (1/4+3/4.x)^4 25/09/2021 By Madeline Hệ số lớn nhất trong khai triển (1/4+3/4.x)^4
Đáp án: \({{27} \over {64}}\) Giải thích các bước giải: $$\eqalign{ & {\left( {{1 \over 4} + {3 \over 4}x} \right)^4} = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 – k}}{{\left( {{3 \over 4}x} \right)}^k}} \cr & = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 – k}}{{\left( {{3 \over 4}} \right)}^k}{x^k}} \cr & Gia\,\,su\,\,{T_{k + 1}} = C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k}\,\,la\,\,he\,\,so\,\,lon\,\,nhat \cr & \Leftrightarrow \left\{ \matrix{ C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k + 1}{\left( {{1 \over 4}} \right)^{3 – k}}{\left( {{3 \over 4}} \right)^{k + 1}} \hfill \cr C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k – 1}{\left( {{1 \over 4}} \right)^{5 – k}}{\left( {{3 \over 4}} \right)^{k – 1}} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {{4!} \over {k!\left( {4 – k} \right)!}}{1 \over 4} > {{4!} \over {\left( {k + 1} \right)!\left( {3 – k} \right)!}}{3 \over 4} \hfill \cr {{4!} \over {k!\left( {4 – k} \right)!}}{3 \over 4} > {{4!} \over {\left( {k – 1} \right)!\left( {5 – k} \right)!}}{1 \over 4} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {1 \over {4 – k}} > {3 \over {k + 1}} \hfill \cr {3 \over k} > {1 \over {5 – k}} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ k + 1 > 12 – 3k \hfill \cr 15 – 3k > k \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 4k > 11 \hfill \cr 4k < 15 \hfill \cr} \right. \cr & \Leftrightarrow {{11} \over 4} < k < {{15} \over 4}.\,\,\,Ma\,\,k \in Z \Rightarrow k = 3\,\,\left( {tm} \right) \cr & Vay\,\,he\,\,so\,\,lon\,\,nhat\,\,la\,\,C_4^3\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}\, = {{27} \over {64}}\, \cr} $$ Trả lời
Đáp án:
\({{27} \over {64}}\)
Giải thích các bước giải:
$$\eqalign{
& {\left( {{1 \over 4} + {3 \over 4}x} \right)^4} = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 – k}}{{\left( {{3 \over 4}x} \right)}^k}} \cr
& = \sum\limits_{k = 0}^4 {C_4^k{{\left( {{1 \over 4}} \right)}^{4 – k}}{{\left( {{3 \over 4}} \right)}^k}{x^k}} \cr
& Gia\,\,su\,\,{T_{k + 1}} = C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k}\,\,la\,\,he\,\,so\,\,lon\,\,nhat \cr
& \Leftrightarrow \left\{ \matrix{
C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k + 1}{\left( {{1 \over 4}} \right)^{3 – k}}{\left( {{3 \over 4}} \right)^{k + 1}} \hfill \cr
C_4^k{\left( {{1 \over 4}} \right)^{4 – k}}{\left( {{3 \over 4}} \right)^k} > C_4^{k – 1}{\left( {{1 \over 4}} \right)^{5 – k}}{\left( {{3 \over 4}} \right)^{k – 1}} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{{4!} \over {k!\left( {4 – k} \right)!}}{1 \over 4} > {{4!} \over {\left( {k + 1} \right)!\left( {3 – k} \right)!}}{3 \over 4} \hfill \cr
{{4!} \over {k!\left( {4 – k} \right)!}}{3 \over 4} > {{4!} \over {\left( {k – 1} \right)!\left( {5 – k} \right)!}}{1 \over 4} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{1 \over {4 – k}} > {3 \over {k + 1}} \hfill \cr
{3 \over k} > {1 \over {5 – k}} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
k + 1 > 12 – 3k \hfill \cr
15 – 3k > k \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
4k > 11 \hfill \cr
4k < 15 \hfill \cr} \right. \cr & \Leftrightarrow {{11} \over 4} < k < {{15} \over 4}.\,\,\,Ma\,\,k \in Z \Rightarrow k = 3\,\,\left( {tm} \right) \cr & Vay\,\,he\,\,so\,\,lon\,\,nhat\,\,la\,\,C_4^3\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}\, = {{27} \over {64}}\, \cr} $$