Đáp án:

107B

108D

Giải thích các bước giải:

\(\begin{array}{l}
Bai\,\,107:\\
{\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{x^k}} \\
 \Rightarrow {T_{k + 1}} = C_n^k{x^k}\\
\frac{{C_n^k}}{{C_n^{k + 1}}} = \frac{{n!}}{{k!\left( {n - k} \right)!}}:\frac{{n!}}{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{n!}}{{k!\left( {n - k} \right)!}}.\frac{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}}{{n!}}\\
\,\,\,\,\,\,\,\,\,\,\, = \frac{{k + 1}}{{n - k}} = \frac{7}{{15}}\\
 \Leftrightarrow 15k + 15 = 7n - 7k\\
 \Leftrightarrow 22k + 15 = 7n\\
Ta\,\,co:\,\,0 \le k \le n\\
 \Rightarrow {n_{\min }} \Leftrightarrow {k_{\min }}\,\,va\,\,22k + 15\,\,\, \vdots \,\,7\\
 \Rightarrow k + 1\,\, \vdots \,\,7 \Rightarrow k = 6.\\
 \Rightarrow {n_{\min }} = \frac{{22.6 + 15}}{7} = 21\\
Chon\,\,B.\\
Bai\,108:\\
{\left( {2x + \frac{1}{{{x^2}}}} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{{\left( {2x} \right)}^{n - k}}{{\left( {{x^{ - 2}}} \right)}^k}}  = \sum\limits_{k = 0}^n {C_n^k{2^{n - k}}{x^{n - 3k}}} \\
{T_{k + 1}} = C_n^k{2^{n - k}}{x^{n - 3k}}\\
{T_3} = C_n^2{.2^{n - 2}}.{x^{n - 6}}\,\,khong\,\,chua\,\,x \Rightarrow n - 6 = 0 \Leftrightarrow n = 6.\\
{\left( {1 + {x^3}} \right)^{30}} = \sum\limits_{l = 0}^{30} {C_{30}^l{x^{3l}}} \\
{T_{l + 1}}' = C_{30}^l{x^{3l}}\\
{T_2}' = C_{30}^1{x^3}\\
{T_3} = {T_2}' \Rightarrow C_6^2{.2^4} = C_{30}^1{x^3}\\
 \Leftrightarrow 240 = 30{x^3} \Leftrightarrow x = 2\\
Chon\,\,D.
\end{array}\)

Đáp án:

107 , n=21

108 , x=2

Giải thích các bước giải:

\(\begin{array}{l}
107,\frac{{C_n^k}}{{C_n^{k + 1}}} = \frac{{\frac{{n!}}{{k!.(n - k)!}}}}{{\frac{{n!}}{{(k + 1)!.(n - k - 1)!}}}}\\
 = \frac{{k + 1}}{{n - k}} = \frac{7}{{15}}\\
 \Rightarrow k = 6,n = 21\\
108,\\
n - k = 2k\\
 \Rightarrow n = 3k\\
k = 2 \Rightarrow n = 6\\
 \Rightarrow C_6^2{.2^4} = C_{30}^1.{({x^3})^1}\\
 \Rightarrow x = 2
\end{array}\)