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Câu hỏi: Cho khai triển nhị thức ${{\left( \dfrac{1}{3}+\dfrac{2}{3}x \right)}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{10}}{{x}^{10}}$. Hệ số ${{a}_{k}}$ lớn nhất trong khai triển trên khi $k$ bằng
A. 5.
B. 3.
C. 6.
D. 7.
A. 5.
B. 3.
C. 6.
D. 7.
Số hạng tổng quát của khai triển: $C_{10}^{k}{{\left( \dfrac{1}{3} \right)}^{10-k}}.{{\left( \dfrac{2}{3}x \right)}^{k}}=C_{10}^{k}\dfrac{{{2}^{k}}}{{{3}^{10}}}{{x}^{k}}={{a}_{k}}.{{x}^{k}}\left( k\ge 0 \right)$.
Giả sử ${{a}_{k}}$ là hệ số lớn nhất thì $\left\{ \begin{aligned}
& {{a}_{k}}\ge {{a}_{k+1}} \\
& {{a}_{k}}\ge {{a}_{k-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& C_{10}^{k}\dfrac{{{2}^{k}}}{{{3}^{10}}}\ge C_{10}^{k+1}\dfrac{{{2}^{k+1}}}{{{3}^{10}}} \\
& C_{10}^{k}\dfrac{{{2}^{k}}}{{{3}^{10}}}\ge C_{10}^{k-1}\dfrac{{{2}^{k-1}}}{{{3}^{10}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{10!}{\left( 10-k \right)!.k!}\ge \dfrac{2.10!}{\left( 9-k \right)!.\left( k+1 \right)!} \\
& \dfrac{2.10!}{\left( 10-k \right)!.k!}\ge \dfrac{10!}{\left( 11-k \right)!.\left( k-1 \right)!} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{10-k}\ge \dfrac{2}{k+1} \\
& \dfrac{2}{k}\ge \dfrac{1}{11-k} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k+1\ge 2\left( 10-k \right) \\
& 2\left( 11-k \right)\ge k \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k\ge \dfrac{19}{3} \\
& k\le \dfrac{22}{3} \\
\end{aligned} \right.\Rightarrow k=7$.
Giả sử ${{a}_{k}}$ là hệ số lớn nhất thì $\left\{ \begin{aligned}
& {{a}_{k}}\ge {{a}_{k+1}} \\
& {{a}_{k}}\ge {{a}_{k-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& C_{10}^{k}\dfrac{{{2}^{k}}}{{{3}^{10}}}\ge C_{10}^{k+1}\dfrac{{{2}^{k+1}}}{{{3}^{10}}} \\
& C_{10}^{k}\dfrac{{{2}^{k}}}{{{3}^{10}}}\ge C_{10}^{k-1}\dfrac{{{2}^{k-1}}}{{{3}^{10}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{10!}{\left( 10-k \right)!.k!}\ge \dfrac{2.10!}{\left( 9-k \right)!.\left( k+1 \right)!} \\
& \dfrac{2.10!}{\left( 10-k \right)!.k!}\ge \dfrac{10!}{\left( 11-k \right)!.\left( k-1 \right)!} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{10-k}\ge \dfrac{2}{k+1} \\
& \dfrac{2}{k}\ge \dfrac{1}{11-k} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k+1\ge 2\left( 10-k \right) \\
& 2\left( 11-k \right)\ge k \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k\ge \dfrac{19}{3} \\
& k\le \dfrac{22}{3} \\
\end{aligned} \right.\Rightarrow k=7$.
Đáp án D.