Tìm hệ số của $x^{3}$ tronng khai triển $\left ( 1+x \right )^{17}\left ( 1-5x \right )$ 
$=(1+x)^{17}-5x(1+x)^{17}=\sum_{k=0}^{17}.C^{k}_{17}.x^{k}-5.\sum_{a=0}^{17}.C^{a}_{17}.x^{a+1}  $
ta phải tìm hệ số của x^{3} ta có hệ \begin{cases}k=3 \\ a+1= 3\end{cases}=> k=3 và a=2. nên hệ số $C^{3}_{17}-5.C^{2}_{17}=0$
Đáp án: 0    la sao bạn
đáp án của phép tính là 0 –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 07-12-14 10:16 PM
sao lại 0 zax? –  nambttvqht 07-12-14 04:57 PM
la sao bạn –  ghost_angle74 07-12-14 04:33 PM

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