Đáp án:
\(C_{10}^{4}.(\frac{2}{3})^{4}\)
Giải thích các bước giải:
Hệ số tổng quát: \(C_{10}^{k}.1^{10-k}.(\frac{2}{3}x)^{k}\)
Ta có:
\(\left\{\begin{matrix} T_{k+1} < T_{k}
& & \\ T_{k-1}<T_{k}
& &
\end{matrix}\right.\)
\( \Leftrightarrow \left\{\begin{matrix} C_{10}^{k+1}.1^{9-k}.(\frac{2x}{3})^{k+1}< C_{10}^{k}.1^{10-k}.(\frac{2x}{3})^{k}
& & \\ C_{10}^{k-1}.1^{11-k}.(\frac{2x}{3})^{k-1}< C_{10}^{k}.1^{10-k}.(\frac{2x}{3})^{k}
& &
\end{matrix}\right.\)
\( \Leftrightarrow \left\{\begin{matrix} \frac{10!}{(9-k)!(k+1)!}.(\frac{2}{3})^{k+1}<\frac{10!}{(10-k)!k!}(\frac{2}{3})^{k}
& & \\ \frac{10!}{(11-k)!(k-1)!}.(\frac{2}{3})^{k-1}<\frac{10!}{(10-k)!k!}.(\frac{2}{3})^{k}
& &
\end{matrix}\right.\)
\( \Leftrightarrow \left\{\begin{matrix} \frac{1}{k+1}.(\frac{2}{3})^{k+1}< \frac{1}{10-k}.(\frac{2}{3})^{k}
& & \\ \frac{1}{11-k}.(\frac{2}{3})^{k-1}< \frac{1}{k}.(\frac{2}{3})^{k}
& &
\end{matrix}\right.\)
\( \Leftrightarrow \left\{\begin{matrix} \frac{1}{k+1}.\frac{2}{3}<\frac{1}{10-k}
& & \\ \frac{1}{11-k}<\frac{1}{k}.\frac{2}{3}
& &
\end{matrix}\right.\)
\( \Leftrightarrow \frac{17}{5}<k<\frac{22}{5}\)
Vậy k=4.
Hệ số lớn nhất:
\(C_{10}^{4}.(\frac{2}{3})^{4}\)
