新高考數(shù)學(xué)三輪沖刺專題14 二項(xiàng)式定理、復(fù)數(shù)（5大易錯(cuò)點(diǎn)分析+解題模板+舉一反三+易錯(cuò)題通關(guān)）（含解析）

資料整理【淘寶店鋪：向陽(yáng)百分百】專題14二項(xiàng)式定理、復(fù)數(shù)易錯(cuò)點(diǎn)一：忽略了二項(xiàng)式中的負(fù)號(hào)而致錯(cuò)（（a-b）n化解問(wèn)題）Ⅰ：二項(xiàng)式定理一般地，對(duì)于任意正整數(shù)，都有：SKIPIF1<0，這個(gè)公式所表示的定理叫做二項(xiàng)式定理，等號(hào)右邊的多項(xiàng)式叫做的二項(xiàng)展開式．式中的SKIPIF1<0做二項(xiàng)展開式的通項(xiàng)，用SKIPIF1<0表示，即通項(xiàng)為展開式的第SKIPIF1<0項(xiàng)：SKIPIF1<0，其中的系數(shù)SKIPIF1<0（r=0，1，2，…，n）叫做二項(xiàng)式系數(shù)，Ⅱ：二項(xiàng)式SKIPIF1<0的展開式的特點(diǎn)：①項(xiàng)數(shù)：共有SKIPIF1<0項(xiàng)，比二項(xiàng)式的次數(shù)大1；②二項(xiàng)式系數(shù)：第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0，最大二項(xiàng)式系數(shù)項(xiàng)居中；③次數(shù)：各項(xiàng)的次數(shù)都等于二項(xiàng)式的冪指數(shù)SKIPIF1<0．字母SKIPIF1<0降冪排列，次數(shù)由SKIPIF1<0到SKIPIF1<0；字母SKIPIF1<0升冪排列，次數(shù)從SKIPIF1<0到SKIPIF1<0，每一項(xiàng)中，SKIPIF1<0，SKIPIF1<0次數(shù)和均為SKIPIF1<0；④項(xiàng)的系數(shù)：二項(xiàng)式系數(shù)依次是SKIPIF1<0，項(xiàng)的系數(shù)是SKIPIF1<0與SKIPIF1<0的系數(shù)（包括二項(xiàng)式系數(shù)）．Ⅲ：兩個(gè)常用的二項(xiàng)展開式：①（）②Ⅳ：二項(xiàng)展開式的通項(xiàng)公式二項(xiàng)展開式的通項(xiàng)：SKIPIF1<0SKIPIF1<0公式特點(diǎn)：①它表示二項(xiàng)展開式的第SKIPIF1<0項(xiàng)，該項(xiàng)的二項(xiàng)式系數(shù)是；②字母SKIPIF1<0的次數(shù)和組合數(shù)的上標(biāo)相同；③SKIPIF1<0與SKIPIF1<0的次數(shù)之和為SKIPIF1<0．注意：①二項(xiàng)式SKIPIF1<0的二項(xiàng)展開式的第r+1項(xiàng)和SKIPIF1<0的二項(xiàng)展開式的第r+1項(xiàng)是有區(qū)別的，應(yīng)用二項(xiàng)式定理時(shí)，其中的SKIPIF1<0和SKIPIF1<0是不能隨便交換位置的．②通項(xiàng)是針對(duì)在SKIPIF1<0這個(gè)標(biāo)準(zhǔn)形式下而言的，如SKIPIF1<0的二項(xiàng)展開式的通項(xiàng)是（只需把SKIPIF1<0看成SKIPIF1<0代入二項(xiàng)式定理）．易錯(cuò)提醒：在二項(xiàng)式定理SKIPIF1<0的問(wèn)題要注意SKIPIF1<0的系數(shù)為SKIPIF1<0，在展開求解時(shí)不要忽略.例、已知SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為30，則SKIPIF1<0（）A．SKIPIF1<0B．SKIPIF1<0C．6D．SKIPIF1<0錯(cuò)解：SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0．錯(cuò)因分析：二項(xiàng)式SKIPIF1<0中的項(xiàng)為SKIPIF1<0，SKIPIF1<0，錯(cuò)解中誤認(rèn)為是SKIPIF1<0，SKIPIF1<0，忽略了負(fù)號(hào)而出現(xiàn)了錯(cuò)解．正解：DSKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0．變式1：在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是．【詳解】二項(xiàng)式SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0（其中SKIPIF1<0且SKIPIF1<0），令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)是SKIPIF1<0.故答案為：SKIPIF1<0變式2：SKIPIF1<0展開式的常數(shù)項(xiàng)為.【詳解】展開式的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以常數(shù)項(xiàng)為SKIPIF1<0，故答案為：15.變式3：SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為.【詳解】設(shè)展開式中的第SKIPIF1<0項(xiàng)含有SKIPIF1<0項(xiàng)，即SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<01．SKIPIF1<0的二項(xiàng)式展開式中SKIPIF1<0的系數(shù)為（

）A．560 B．35 C．-35 D．-560【答案】D【分析】SKIPIF1<0中利用二項(xiàng)式定理可求得SKIPIF1<0的系數(shù)，從而求解.【詳解】由題意知SKIPIF1<0的展開式為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0，故D項(xiàng)正確.故選：D.2．若SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為16，則SKIPIF1<0的展開式中的常數(shù)項(xiàng)為（

）A．6 B．8 C．28 D．56【答案】C【分析】根據(jù)SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和求出n的值，從而寫出SKIPIF1<0的展開式的通項(xiàng)公式，再令x的指數(shù)為0，即可求解常數(shù)項(xiàng)．【詳解】由SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為16，得SKIPIF1<0，所以SKIPIF1<0，則二項(xiàng)式SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的展開式中的常數(shù)項(xiàng)為28，故選：C．3．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為（

）A．55 B．SKIPIF1<0 C．65 D．SKIPIF1<0【答案】D【分析】根據(jù)SKIPIF1<0展開式的通項(xiàng)公式進(jìn)行計(jì)算即可.【詳解】含SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0．故選：SKIPIF1<04．若SKIPIF1<0的展開式中含有常數(shù)項(xiàng)（非零），則正整數(shù)SKIPIF1<0的可能值是（

）A．3 B．4 C．5 D．6【答案】C【分析】根據(jù)二項(xiàng)展開式的通項(xiàng)公式建立方程，求解即可.【詳解】由二項(xiàng)式定理知，SKIPIF1<0SKIPIF1<0，因?yàn)槠浜谐?shù)項(xiàng)，即存在SKIPIF1<0，使得SKIPIF1<0此時(shí)SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，故選：C.5．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0（

）A．2 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用二項(xiàng)式的展開式公式展開，再與前面的項(xiàng)相乘求解即可.【詳解】SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，所以SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0．由題意，可知SKIPIF1<0，所以SKIPIF1<0．故選：D．6．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（

）A．SKIPIF1<0 B．21 C．189 D．SKIPIF1<0【答案】B【分析】利用二項(xiàng)展開式的通項(xiàng)公式可得解.【詳解】由二項(xiàng)展開式的通項(xiàng)公式得SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：B.7．SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為.【答案】960【分析】利用二項(xiàng)展開式的通項(xiàng)公式分析運(yùn)算求解.【詳解】SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0，故令SKIPIF1<0，可得SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為：SKIPIF1<0.故答案為：960.8．已知SKIPIF1<0的展開式中的常數(shù)項(xiàng)是672，則SKIPIF1<0.【答案】2【分析】寫出二項(xiàng)式通項(xiàng)SKIPIF1<0，整理后讓x的次數(shù)為0，得出r的值，再根據(jù)常數(shù)項(xiàng)的值列出等式方程即可得出a的值．【詳解】展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以常數(shù)項(xiàng)是SKIPIF1<0，故SKIPIF1<0．故答案為：2．9．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為．【答案】24【分析】求出二項(xiàng)式展開式的通項(xiàng)公式，再求出指定項(xiàng)的系數(shù)即得.【詳解】二項(xiàng)式SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，所以x的系數(shù)為24.故答案為：24.10．SKIPIF1<0的展開式中，按SKIPIF1<0的升冪排列的第3項(xiàng)的系數(shù)為．【答案】3【分析】根據(jù)已知得出按SKIPIF1<0的升冪排列的第3項(xiàng)即含SKIPIF1<0的項(xiàng).結(jié)合二項(xiàng)式定理，分類討論求解，即可得出答案.【詳解】由已知可得，展開式中含有常數(shù)項(xiàng)、一次項(xiàng)、兩次項(xiàng)，所以，按SKIPIF1<0的升冪排列的第3項(xiàng)即含SKIPIF1<0的項(xiàng).SKIPIF1<0展開式中的常數(shù)項(xiàng)為SKIPIF1<0，SKIPIF1<0展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0；SKIPIF1<0展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，SKIPIF1<0展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0；SKIPIF1<0展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，SKIPIF1<0展開式中的常數(shù)項(xiàng)為SKIPIF1<0.所以，SKIPIF1<0的展開式中，含SKIPIF1<0的項(xiàng)為SKIPIF1<0.故答案為：3.11．在SKIPIF1<0的展開式中的SKIPIF1<0的系數(shù)是.【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)展開式的通項(xiàng)公式，可令SKIPIF1<0求得SKIPIF1<0的系數(shù).【詳解】SKIPIF1<0展開式的通項(xiàng)公式為：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.12．二項(xiàng)式SKIPIF1<0的展開式中常數(shù)項(xiàng)為．【答案】SKIPIF1<0【分析】根據(jù)給定的條件，利用二項(xiàng)式定理求解作答.【詳解】SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0．令SKIPIF1<0，得SKIPIF1<0，故常數(shù)項(xiàng)為SKIPIF1<0．故答案為：SKIPIF1<0.13．SKIPIF1<0的展開式的第三項(xiàng)的系數(shù)為135，則SKIPIF1<0．【答案】6【分析】先寫出展開式的通項(xiàng)公式SKIPIF1<0；再令SKIPIF1<0，列出等式求解即可.【詳解】SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，則第三項(xiàng)的系數(shù)為SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0（舍去）或SKIPIF1<0．故答案為：6.易錯(cuò)點(diǎn)二：三項(xiàng)式轉(zhuǎn)化不合理導(dǎo)致計(jì)算麻煩失誤（三項(xiàng)展開式的問(wèn)題）求三項(xiàng)展開式式中某些特定項(xiàng)的系數(shù)的方法第一步：通過(guò)變形先把三項(xiàng)式轉(zhuǎn)化為二項(xiàng)式，再用二項(xiàng)式定理求解第二步：兩次利用二項(xiàng)式定理的通項(xiàng)公式求解第三步：由二項(xiàng)式定理的推證方法知，可用排列、組合的基本原理去求，即把三項(xiàng)式看作幾個(gè)因式之積，要得到特定項(xiàng)看有多少種方法從這幾個(gè)因式中取因式中的量易錯(cuò)提醒：對(duì)于三項(xiàng)式的展開問(wèn)題，一般采取轉(zhuǎn)化為二項(xiàng)式再展開的辦法進(jìn)行求解，但在轉(zhuǎn)化為二項(xiàng)式的時(shí)候，又有不同的處理策略：一是如果三項(xiàng)式能夠化為完全平方的形式，或者能夠進(jìn)行因式分解，則可通過(guò)對(duì)分解出來(lái)的兩個(gè)二項(xiàng)展開式分別進(jìn)行分析，進(jìn)而解決問(wèn)題（如本例中的解法二）；二是不能化為完全平方的形式，也不能進(jìn)行因式分解時(shí)，可直接將三項(xiàng)式加括號(hào)變?yōu)槎?xiàng)式，套用通項(xiàng)公式展開后對(duì)其中的二項(xiàng)式再利用通項(xiàng)展開并進(jìn)行分析求解，但要結(jié)合要求解的問(wèn)題進(jìn)行合理的變形，以利于求解.例、SKIPIF1<0的展開式中，x的一次項(xiàng)的系數(shù)為（）A．120B．240C．320D．480易錯(cuò)分析：本題易出現(xiàn)的錯(cuò)誤是盲目套用解決三項(xiàng)式展開的一般方法（轉(zhuǎn)化為二項(xiàng)式處理：SKIPIF1<0），而不針對(duì)要求解的問(wèn)題進(jìn)行合理的變通，導(dǎo)致運(yùn)算繁雜并出現(xiàn)錯(cuò)誤．正解：解法一由于SKIPIF1<0，展開式的通項(xiàng)為SKIPIF1<0，0≤r≤5，當(dāng)且僅當(dāng)r＝1時(shí)，展開式才有x的一次項(xiàng)，此時(shí)SKIPIF1<0．所以展開式中x的一次項(xiàng)為SKIPIF1<0，它的系數(shù)為SKIPIF1<0．故選B．解法二由于SKIPIF1<0，所以展開式中x的一次項(xiàng)為SKIPIF1<0．故x的一次項(xiàng)的系數(shù)為240．故選B．變式1：在SKIPIF1<0的展開式中，含SKIPIF1<0的系數(shù)為.【詳解】把SKIPIF1<0的展開式看成是5個(gè)因式SKIPIF1<0的乘積形式，展開式中，含SKIPIF1<0項(xiàng)的系數(shù)可以按如下步驟得到：第一步，從5個(gè)因式中任選2個(gè)因式，這2個(gè)因式取SKIPIF1<0，有SKIPIF1<0種取法；第二步，從剩余的3個(gè)因式中任選2個(gè)因式，都取SKIPIF1<0，有SKIPIF1<0種取法；第三步，把剩余的1個(gè)因式中取SKIPIF1<0，有SKIPIF1<0種取法；根據(jù)分步相乘原理，得；含SKIPIF1<0項(xiàng)的系數(shù)是SKIPIF1<0故答案為：SKIPIF1<0.變式2：SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為（用數(shù)字作答）．【詳解】由于SKIPIF1<0表示5個(gè)因式SKIPIF1<0的乘積，故其中有2個(gè)因式取SKIPIF1<0，2個(gè)因式取SKIPIF1<0，剩余的一個(gè)因式取SKIPIF1<0，可得含SKIPIF1<0的項(xiàng)，故展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，故答案為：SKIPIF1<0．變式3：在SKIPIF1<0的展開式中，形如SKIPIF1<0的所有項(xiàng)系數(shù)之和是．【詳解】SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0．令SKIPIF1<0，得SKIPIF1<0．令SKIPIF1<0，得所求系數(shù)之和為SKIPIF1<0．故答案為：3201．SKIPIF1<0的展開式中的常數(shù)項(xiàng)為（

）A．588 B．589 C．798 D．799【答案】B【分析】因?yàn)镾KIPIF1<0展開式中的項(xiàng)可以看作8個(gè)含有三個(gè)單項(xiàng)式SKIPIF1<0各取一個(gè)相乘而得，分析組合可能，結(jié)合組合數(shù)運(yùn)算求解.【詳解】因?yàn)镾KIPIF1<0展開式中的項(xiàng)可以看作8個(gè)含有三個(gè)單項(xiàng)式SKIPIF1<0中各取一個(gè)相乘而得，若得到常數(shù)項(xiàng)，則有：①8個(gè)1；②2個(gè)SKIPIF1<0，1個(gè)SKIPIF1<0，5個(gè)1；③4個(gè)SKIPIF1<0，2個(gè)SKIPIF1<0，2個(gè)1；所以展開式中的常數(shù)項(xiàng)為SKIPIF1<0.故選：B.2．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是（

）A．24 B．32 C．36 D．40【答案】D【分析】根據(jù)題意，SKIPIF1<0的項(xiàng)為SKIPIF1<0，化簡(jiǎn)后即可求解.【詳解】根據(jù)題意，SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的系數(shù)是SKIPIF1<0.故選：D．3．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為12，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)乘法的運(yùn)算法則，結(jié)合組合數(shù)的性質(zhì)、二倍角的余弦公式進(jìn)行求解即可.【詳解】SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)可以看成：6個(gè)因式SKIPIF1<0中選取5個(gè)因式提供SKIPIF1<0，余下一個(gè)因式中提供SKIPIF1<0或者6個(gè)因式SKIPIF1<0中選取4個(gè)因式提供SKIPIF1<0，余下兩個(gè)因式中均提供SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故選：C4．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為（

）A．SKIPIF1<0 B．60 C．SKIPIF1<0 D．120【答案】A【分析】先把SKIPIF1<0看作整體寫出二項(xiàng)式展開的通項(xiàng)，再根據(jù)指定項(xiàng)確定SKIPIF1<0的次數(shù)，再寫一次二項(xiàng)式展開的通項(xiàng)，最后根據(jù)指定項(xiàng)配湊出項(xiàng)的系數(shù).【詳解】因?yàn)镾KIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)才能出現(xiàn)SKIPIF1<0，此時(shí)SKIPIF1<0展開的通項(xiàng)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)出現(xiàn)SKIPIF1<0的一次，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0．故選：A.5．設(shè)SKIPIF1<0，已知SKIPIF1<0的展開式中只有第5項(xiàng)的二項(xiàng)式系數(shù)最大，且展開式中所有項(xiàng)的系數(shù)和為256，則SKIPIF1<0中SKIPIF1<0的系數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)題意得到SKIPIF1<0和SKIPIF1<0，再根據(jù)SKIPIF1<0項(xiàng)的取法為1個(gè)SKIPIF1<0和1個(gè)SKIPIF1<0再計(jì)算即可.【詳解】因?yàn)镾KIPIF1<0的展開式中只有第5項(xiàng)的二項(xiàng)式系數(shù)最大，所以展開式一共有SKIPIF1<0項(xiàng)，即SKIPIF1<0，令SKIPIF1<0，得展開式中所有項(xiàng)的系數(shù)和為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0中SKIPIF1<0項(xiàng)的取法為1個(gè)SKIPIF1<0和1個(gè)SKIPIF1<0，所以SKIPIF1<0系數(shù)為SKIPIF1<0.故選：C6．SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（

）A．80 B．60 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由題得SKIPIF1<0，再利用二項(xiàng)式的通項(xiàng)即可得到答案.【詳解】SKIPIF1<0，則其展開式通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0，故選：D．7．已知SKIPIF1<0展開式的各項(xiàng)系數(shù)之和為SKIPIF1<0，則展開式中SKIPIF1<0的系數(shù)為（

）A．270 B．SKIPIF1<0 C．330 D．SKIPIF1<0【答案】D【分析】令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0.再根據(jù)二項(xiàng)展開式的通項(xiàng)公式即可求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，又因?yàn)橹挥蠸KIPIF1<0，SKIPIF1<0展開式中有含SKIPIF1<0的項(xiàng)，所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：D8．SKIPIF1<0的展開式中只有第5項(xiàng)的二項(xiàng)式系數(shù)最大，若展開式中所有項(xiàng)的系數(shù)和為256，則SKIPIF1<0中SKIPIF1<0的系數(shù)為（

）A．1 B．4或1 C．4或0 D．6或0【答案】C【分析】展開式中只有第5項(xiàng)的二項(xiàng)式系數(shù)最大，可以得到SKIPIF1<0的值，然后再賦值法求出所有項(xiàng)的系數(shù)和的表達(dá)式可解出a的值，再分類求出SKIPIF1<0中SKIPIF1<0的系數(shù)即可得出答案.【詳解】展開式中只有第5項(xiàng)的二項(xiàng)式系數(shù)最大，所以總共有9項(xiàng)，SKIPIF1<0令SKIPIF1<0得所有項(xiàng)的系數(shù)和為SKIPIF1<0，SKIPIF1<0或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0展開式中不含SKIPIF1<0項(xiàng).故選：C.9．SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)的系數(shù)為．【答案】80【分析】只需6個(gè)因式中3個(gè)因式取SKIPIF1<0、3個(gè)因式取SKIPIF1<0或2個(gè)因式取SKIPIF1<0、1個(gè)因式取SKIPIF1<0、3個(gè)因式取1，根據(jù)組合知識(shí)得到答案.【詳解】SKIPIF1<0可以看成6個(gè)因式SKIPIF1<0相乘，所以SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為3個(gè)因式取SKIPIF1<0、3個(gè)因式取SKIPIF1<0或2個(gè)因式取SKIPIF1<0、1個(gè)因式取SKIPIF1<0、3個(gè)因式取1，所以SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0．故答案為：8010．SKIPIF1<0展開式中，SKIPIF1<0項(xiàng)的系數(shù)為.【答案】SKIPIF1<0【分析】由二項(xiàng)式定理求解．【詳解】SKIPIF1<0，∵SKIPIF1<0的指數(shù)是3，∴得到SKIPIF1<0，∵SKIPIF1<0的指數(shù)是2，得到SKIPIF1<0，∴SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<011．SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)的系數(shù)為．【答案】SKIPIF1<0【分析】根據(jù)多項(xiàng)式相乘展開方法求解.【詳解】SKIPIF1<0的展開式中，構(gòu)成SKIPIF1<0項(xiàng)只能是一個(gè)SKIPIF1<0、一個(gè)SKIPIF1<0、3個(gè)SKIPIF1<0相乘，故此項(xiàng)為SKIPIF1<0．故答案為：SKIPIF1<0.12．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為.【答案】66【分析】根據(jù)二項(xiàng)式的含義，結(jié)合組合數(shù)的計(jì)算，求得答案.【詳解】由題意，SKIPIF1<0表示12個(gè)因式SKIPIF1<0的乘積，故當(dāng)2個(gè)因式取x，其余10個(gè)因式取1時(shí)，可得展開式中含SKIPIF1<0的項(xiàng)，故SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：66.13．SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為10，則SKIPIF1<0.【答案】SKIPIF1<0【分析】化SKIPIF1<0,利用二項(xiàng)展開式的通項(xiàng)公式求得展開式中SKIPIF1<0的系數(shù),列方程求出SKIPIF1<0的值.【詳解】SKIPIF1<0其展開式的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0因?yàn)镾KIPIF1<0的系數(shù)為10，則SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0.14．SKIPIF1<0展開式中的常數(shù)項(xiàng)為.（用數(shù)字做答）【答案】49【分析】利用分類計(jì)數(shù)原理求解即可.【詳解】SKIPIF1<0展開式中得到常數(shù)項(xiàng)的方法分類如下：（1）4個(gè)因式中都不取SKIPIF1<0，則不取SKIPIF1<0，全取SKIPIF1<0，相乘得到常數(shù)項(xiàng).常數(shù)項(xiàng)為SKIPIF1<0；（2）4個(gè)因式中有1個(gè)取SKIPIF1<0，則再取1個(gè)SKIPIF1<0，其余因式取SKIPIF1<0，相乘得到常數(shù)項(xiàng).常數(shù)項(xiàng)為SKIPIF1<0；（3）4個(gè)因式中有2個(gè)取SKIPIF1<0，則再取2個(gè)SKIPIF1<0，相乘得到常數(shù)項(xiàng).常數(shù)項(xiàng)為SKIPIF1<0.合并同類項(xiàng)，所以展開式中常數(shù)項(xiàng)為SKIPIF1<0.故答案為：SKIPIF1<0.15．SKIPIF1<0展開式中含SKIPIF1<0項(xiàng)的系數(shù)為．【答案】-160【分析】變形為SKIPIF1<0，寫出通項(xiàng)公式，求出SKIPIF1<0，得到答案.【詳解】SKIPIF1<0變形為SKIPIF1<0，故通項(xiàng)公式得SKIPIF1<0，其中SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，故通項(xiàng)公式為SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0.故答案為：-16016．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為．【答案】92【分析】由于SKIPIF1<0，根據(jù)二項(xiàng)式定理分別求得SKIPIF1<0和SKIPIF1<0的展開式的通項(xiàng)，從而分析可得SKIPIF1<0的系數(shù).【詳解】SKIPIF1<0，又SKIPIF1<0展開式的通項(xiàng)SKIPIF1<0，SKIPIF1<0展開式的通項(xiàng)SKIPIF1<0，所以含SKIPIF1<0的項(xiàng)為SKIPIF1<0則含SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.17．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為（用數(shù)字作答）．【答案】SKIPIF1<0【分析】SKIPIF1<0，然后兩次利用通項(xiàng)公式求解即可；【詳解】因?yàn)镾KIPIF1<0，設(shè)其展開式的通項(xiàng)公式為：SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為SKIPIF1<0，故答案為：SKIPIF1<0易錯(cuò)點(diǎn)三：混淆項(xiàng)的系數(shù)與二項(xiàng)式系數(shù)致誤（系數(shù)與二項(xiàng)式系數(shù)問(wèn)題）Ⅰ：二項(xiàng)式展開式中的最值問(wèn)題1.二項(xiàng)式系數(shù)的性質(zhì)=1\*GB3①每一行兩端都是SKIPIF1<0，即SKIPIF1<0；其余每個(gè)數(shù)都等于它“肩上”兩個(gè)數(shù)的和，即SKIPIF1<0．=2\*GB3②對(duì)稱性每一行中，與首末兩端“等距離”的兩個(gè)二項(xiàng)式系數(shù)相等，即SKIPIF1<0．=3\*GB3③二項(xiàng)式系數(shù)和令SKIPIF1<0，則二項(xiàng)式系數(shù)的和為SKIPIF1<0，變形式SKIPIF1<0．=4\*GB3④奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和等于偶數(shù)項(xiàng)的二項(xiàng)式系數(shù)和在二項(xiàng)式定理中，令SKIPIF1<0，則SKIPIF1<0，從而得到：SKIPIF1<0．=5\*GB3⑤最大值：如果二項(xiàng)式的冪指數(shù)SKIPIF1<0是偶數(shù)，則中間一項(xiàng)SKIPIF1<0的二項(xiàng)式系數(shù)SKIPIF1<0最大；如果二項(xiàng)式的冪指數(shù)SKIPIF1<0是奇數(shù)，則中間兩項(xiàng)SKIPIF1<0，SKIPIF1<0的二項(xiàng)式系數(shù)SKIPIF1<0，SKIPIF1<0相等且最大．2.系數(shù)的最大項(xiàng)求SKIPIF1<0展開式中最大的項(xiàng)，一般采用待定系數(shù)法．設(shè)展開式中各項(xiàng)系數(shù)分別為SKIPIF1<0，設(shè)第SKIPIF1<0項(xiàng)系數(shù)最大，應(yīng)有SKIPIF1<0，從而解出SKIPIF1<0來(lái)．Ⅱ：二項(xiàng)式展開式中系數(shù)和有關(guān)問(wèn)題常用賦值舉例：（1）設(shè)，二項(xiàng)式定理是一個(gè)恒等式，即對(duì)SKIPIF1<0，SKIPIF1<0的一切值都成立，我們可以根據(jù)具體問(wèn)題的需要靈活選取SKIPIF1<0，SKIPIF1<0的值．①令，可得：②令SKIPIF1<0，可得：，即：（假設(shè)為偶數(shù)），再結(jié)合①可得：．（2）若SKIPIF1<0，則①常數(shù)項(xiàng)：令SKIPIF1<0，得SKIPIF1<0．②各項(xiàng)系數(shù)和：令SKIPIF1<0，得SKIPIF1<0．注意：常見的賦值為令SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，然后通過(guò)加減運(yùn)算即可得到相應(yīng)的結(jié)果．易錯(cuò)提醒：二項(xiàng)式定理SKIPIF1<0的問(wèn)題要注意：項(xiàng)的系數(shù)與二項(xiàng)式系數(shù)的區(qū)別與聯(lián)系（求所有項(xiàng)的系數(shù)只要令字母值為1）.例、設(shè)SKIPIF1<0的展開式中，第三項(xiàng)的系數(shù)為36，試求含SKIPIF1<0的項(xiàng)．錯(cuò)解：第三項(xiàng)的系數(shù)為SKIPIF1<0，依題意得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，解此方程并舍去不合題意的負(fù)值，得n＝9，設(shè)SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)為第r＋1項(xiàng)，則SKIPIF1<0，由9－r＝2，得r＝7，故SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0．錯(cuò)因分析：錯(cuò)解將“二項(xiàng)展開式中的第三項(xiàng)的二項(xiàng)式系數(shù)”當(dāng)作了“第三項(xiàng)的系數(shù)”，解答顯然是錯(cuò)誤的．正解：SKIPIF1<0的展開式的第三項(xiàng)為SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，解此方程并舍去不合題意的負(fù)值，得n＝4，設(shè)SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)為第r＋1項(xiàng)，則SKIPIF1<0，由4－r＝2，得r＝2，即SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)為SKIPIF1<0．變式1：求SKIPIF1<0的展開式中第3項(xiàng)的系數(shù)和二項(xiàng)式系數(shù)．【詳解】二項(xiàng)式SKIPIF1<0展開式通項(xiàng)公式為SKIPIF1<0，第三項(xiàng)為：SKIPIF1<0，所以第三項(xiàng)系數(shù)為SKIPIF1<0，第3項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0．變式2：計(jì)算SKIPIF1<0的展開式中第5項(xiàng)的系數(shù)和二項(xiàng)式系數(shù)．【詳解】因?yàn)镾KIPIF1<0的展開通項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的展開式中第5項(xiàng)是SKIPIF1<0，故所求第5項(xiàng)的系數(shù)是SKIPIF1<0，第5項(xiàng)的二項(xiàng)式系數(shù)是SKIPIF1<0．變式3：求SKIPIF1<0的展開式中常數(shù)項(xiàng)的值和對(duì)應(yīng)的二項(xiàng)式系數(shù)．【詳解】因?yàn)镾KIPIF1<0，所以展開式中的第SKIPIF1<0項(xiàng)為SKIPIF1<0．要得到常數(shù)項(xiàng)，必須有SKIPIF1<0，從而有SKIPIF1<0，因此常數(shù)項(xiàng)是第4項(xiàng)，且SKIPIF1<0．從而可知常數(shù)項(xiàng)的值為160，其對(duì)應(yīng)的二項(xiàng)式系數(shù)為SKIPIF1<0．1．在二項(xiàng)式SKIPIF1<0的展開式中，二項(xiàng)式系數(shù)最大的是（

）A．第3項(xiàng) B．第4項(xiàng)C．第5項(xiàng) D．第3項(xiàng)和第4項(xiàng)【答案】B【分析】根據(jù)二項(xiàng)式系數(shù)的性質(zhì)分析求解.【詳解】二項(xiàng)式SKIPIF1<0的展開式共有7項(xiàng)，則二項(xiàng)式系數(shù)最大的是第4項(xiàng).故選：B.2．已知二項(xiàng)式SKIPIF1<0的展開式中僅有第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)最大，則SKIPIF1<0為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】分析可知，二項(xiàng)式SKIPIF1<0的展開式共SKIPIF1<0項(xiàng)，即可求出SKIPIF1<0的值.【詳解】因?yàn)槎?xiàng)式SKIPIF1<0的展開式中僅有第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)最大，則二項(xiàng)式SKIPIF1<0的展開式共SKIPIF1<0項(xiàng)，即SKIPIF1<0，解得SKIPIF1<0.故選：A.3．在二項(xiàng)式SKIPIF1<0的展開式中，下列說(shuō)法正確的是（

）A．常數(shù)項(xiàng)是SKIPIF1<0 B．各項(xiàng)系數(shù)和為SKIPIF1<0C．第5項(xiàng)二項(xiàng)式系數(shù)最大 D．奇數(shù)項(xiàng)二項(xiàng)式系數(shù)和為32【答案】BD【分析】根據(jù)二項(xiàng)式理及二項(xiàng)式系數(shù)的性質(zhì)逐項(xiàng)判斷即可.【詳解】二項(xiàng)式SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，得常數(shù)項(xiàng)為SKIPIF1<0，故A不正確；當(dāng)SKIPIF1<0時(shí)，可得展開式各項(xiàng)系數(shù)和為SKIPIF1<0，故B正確；由于SKIPIF1<0，則二項(xiàng)式系數(shù)最大為SKIPIF1<0為展開式的第4項(xiàng)，故C不正確；奇數(shù)項(xiàng)二項(xiàng)式系數(shù)和為SKIPIF1<0，故D正確.故選：BD.4．在二項(xiàng)式SKIPIF1<0的展開式中，下列說(shuō)法正確的是（

）A．第6項(xiàng)的二項(xiàng)式系數(shù)最大 B．第6項(xiàng)的系數(shù)最大C．所有項(xiàng)的二項(xiàng)式系數(shù)之和為SKIPIF1<0 D．所有項(xiàng)的系數(shù)之和為1【答案】ACD【分析】由系數(shù)和二項(xiàng)式的系數(shù)的性質(zhì)可判斷A，B，C；由賦值可判斷D.【詳解】通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0，其二項(xiàng)式系數(shù)為SKIPIF1<0，二項(xiàng)式SKIPIF1<0的展開式共SKIPIF1<0項(xiàng)，中間項(xiàng)的二項(xiàng)式系數(shù)最大，故第6項(xiàng)的二項(xiàng)式系數(shù)SKIPIF1<0是最大的，故A正確；二項(xiàng)式系數(shù)和為SKIPIF1<0，所以C正確；令SKIPIF1<0得所有項(xiàng)的系數(shù)和為1，故D正確；因?yàn)檎归_式中第六項(xiàng)的系數(shù)為負(fù)數(shù)，所以第六項(xiàng)的系數(shù)不可能為最大，故B選項(xiàng)錯(cuò)誤，故選：ACD.5．已知2，n，8成等差數(shù)列，則在SKIPIF1<0的展開式中，下列說(shuō)法正確的是（

）A．二項(xiàng)式系數(shù)之和為32 B．各項(xiàng)系數(shù)之和為1C．常數(shù)項(xiàng)為40 D．展開式中系數(shù)最大的項(xiàng)為80x【答案】ABD【分析】根據(jù)等差中項(xiàng)可得SKIPIF1<0.對(duì)于A：根據(jù)二項(xiàng)式系數(shù)之和的結(jié)論直接運(yùn)算求解；對(duì)于B：利用賦值法運(yùn)算求解；對(duì)于C、D：利用二項(xiàng)展開式的通項(xiàng)公式運(yùn)算求解.【詳解】由題意可得：SKIPIF1<0，則SKIPIF1<0，對(duì)于選項(xiàng)A：二項(xiàng)式系數(shù)之和為SKIPIF1<0，故A正確；對(duì)于選項(xiàng)B：令SKIPIF1<0，可得各項(xiàng)系數(shù)之和為SKIPIF1<0，故B正確；對(duì)于選項(xiàng)C、D：因?yàn)镾KIPIF1<0的展開式的通項(xiàng)公式為：SKIPIF1<0，所以SKIPIF1<0，展開式中沒有常數(shù)項(xiàng)，故C錯(cuò)誤；展開式中系數(shù)最大的項(xiàng)為80x，故D正確；故選：ABD.6．下列關(guān)于SKIPIF1<0的展開式的說(shuō)法中正確的是（

）A．常數(shù)項(xiàng)為－160B．第4項(xiàng)的系數(shù)最大C．第4項(xiàng)的二項(xiàng)式系數(shù)最大D．所有項(xiàng)的系數(shù)和為1【答案】ACD【分析】利用二項(xiàng)展開式的通項(xiàng)和二項(xiàng)式系數(shù)的性質(zhì)求解.【詳解】SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0.對(duì)于A，令SKIPIF1<0，解得SKIPIF1<0，∴常數(shù)項(xiàng)為SKIPIF1<0，A正確；對(duì)于B，由通項(xiàng)公式知，若要系數(shù)最大，k所有可能的取值為0，2，4，6，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴展開式第5項(xiàng)的系數(shù)最大，B錯(cuò)誤；對(duì)于C，展開式共有7項(xiàng)，得第4項(xiàng)的二項(xiàng)式系數(shù)最大，C正確；對(duì)于D，令x＝1，則所有項(xiàng)的系數(shù)和為SKIPIF1<0，D正確.故選：ACD.7．若SKIPIF1<0的展開式的二項(xiàng)式系數(shù)之和為16，則SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為.【答案】56【分析】通過(guò)二項(xiàng)式系數(shù)和求出SKIPIF1<0，然后求出SKIPIF1<0展開式的通項(xiàng)公式，最后求出指定項(xiàng)的系數(shù)即可.【詳解】由SKIPIF1<0的展開式的二項(xiàng)式系數(shù)之和為16，得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：568．已知常數(shù)SKIPIF1<0，在SKIPIF1<0的二項(xiàng)展開式中的常數(shù)項(xiàng)為15，設(shè)SKIPIF1<0，則SKIPIF1<0．【答案】-31【分析】先求出SKIPIF1<0，再由二項(xiàng)式的展開式進(jìn)行求解即可.【詳解】解：SKIPIF1<0的展開式為：SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0的展開式為：SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故答案為：-31.9．在SKIPIF1<0的二項(xiàng)式中，所有的二項(xiàng)式系數(shù)之和為64，則各項(xiàng)的系數(shù)的絕對(duì)值之和為．【答案】729/SKIPIF1<0【分析】根據(jù)二項(xiàng)式系數(shù)之和求出n的值，進(jìn)而設(shè)出各項(xiàng)的系數(shù)，然后采用賦值法即可求得答案.【詳解】由題意SKIPIF1<0的二項(xiàng)式中，所有的二項(xiàng)式系數(shù)之和為64，即SKIPIF1<0，設(shè)SKIPIF1<0的各項(xiàng)的系數(shù)為SKIPIF1<0，則各項(xiàng)的系數(shù)的絕對(duì)值之和為SKIPIF1<0，即為SKIPIF1<0中各項(xiàng)的系數(shù)的和，令SKIPIF1<0，SKIPIF1<0，即各項(xiàng)的系數(shù)的絕對(duì)值之和為SKIPIF1<0，故答案為：72910．二項(xiàng)式SKIPIF1<0的展開式中常數(shù)項(xiàng)為（用數(shù)字作答）.【答案】60【分析】根據(jù)二項(xiàng)式展開式的通項(xiàng)公式即可求得正確答案.【詳解】二項(xiàng)式SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0，由題意令SKIPIF1<0，解得SKIPIF1<0，所以二項(xiàng)式展開式中的常數(shù)項(xiàng)為SKIPIF1<0.故答案為：60.11．已知SKIPIF1<0的展開式中第9項(xiàng)、第10項(xiàng)、第11項(xiàng)的二項(xiàng)式系數(shù)成等差數(shù)列，則SKIPIF1<0.【答案】14或23【分析】根據(jù)二項(xiàng)式系數(shù)的定義列出等式，解方程即可求得SKIPIF1<0或SKIPIF1<0.【詳解】由題意可得SKIPIF1<0成等差數(shù)列，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：14或2312．SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為.【答案】SKIPIF1<0【分析】先對(duì)第一個(gè)括號(hào)中選取單項(xiàng)式進(jìn)行分類，然后再在每一類中分步，結(jié)合計(jì)數(shù)原理以及組合數(shù)即可求解.【詳解】要得到SKIPIF1<0的展開式中含有SKIPIF1<0的項(xiàng)，分以下兩種情形：情形一：先在第一個(gè)括號(hào)中選取“SKIPIF1<0”，然后在后面四個(gè)括號(hào)中選取3個(gè)“SKIPIF1<0”和1個(gè)“SKIPIF1<0”，由分步乘法計(jì)數(shù)原理可知此時(shí)“SKIPIF1<0”的系數(shù)為SKIPIF1<0；情形二：先在第一個(gè)括號(hào)中選取“SKIPIF1<0”，然后在后面四個(gè)括號(hào)中選取2個(gè)“SKIPIF1<0”和2個(gè)“SKIPIF1<0”，由分步乘法計(jì)數(shù)原理可知此時(shí)“SKIPIF1<0”的系數(shù)為SKIPIF1<0.綜上所述：由分類加法計(jì)數(shù)原理可知SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.13．若SKIPIF1<0展開式的二項(xiàng)式系數(shù)和為64，則展開式中第三項(xiàng)的二項(xiàng)式系數(shù)為.【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)式系數(shù)和得到SKIPIF1<0，再計(jì)算第三項(xiàng)的二項(xiàng)式系數(shù)即可.【詳解】SKIPIF1<0展開式的二項(xiàng)式系數(shù)和為SKIPIF1<0，故SKIPIF1<0，展開式中第三項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.14．若SKIPIF1<0的展開式中二項(xiàng)式系數(shù)之和為64，則展開式中的常數(shù)項(xiàng)是.【答案】SKIPIF1<0【分析】先求得SKIPIF1<0的值，然后根據(jù)二項(xiàng)式展開式的通項(xiàng)公式求得正確答案.【詳解】依題意，SKIPIF1<0，則二項(xiàng)式SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以展開式中的常數(shù)項(xiàng)是SKIPIF1<0.故答案為：SKIPIF1<015．已知SKIPIF1<0，若SKIPIF1<0展開式各項(xiàng)的二項(xiàng)式系數(shù)的和為1024，則SKIPIF1<0的值為．【答案】17010【分析】由題意，利用二項(xiàng)式系數(shù)的性質(zhì)求出SKIPIF1<0值，再根據(jù)二項(xiàng)式展開式的通項(xiàng)公式，求出SKIPIF1<0值．【詳解】SKIPIF1<0SKIPIF1<0，SKIPIF1<0展開式各項(xiàng)的二項(xiàng)式系數(shù)的和為SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0．則令SKIPIF1<0，可得SKIPIF1<0．故答案為：17010．16．已知SKIPIF1<0的展開式中二項(xiàng)式系數(shù)和是64，則展開式中x的系數(shù)為．【答案】60【分析】手續(xù)愛你根據(jù)二項(xiàng)式系數(shù)和公式求出SKIPIF1<0，再利用二項(xiàng)展開式的通項(xiàng)公式即可得到答案.【詳解】由題意得SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的二項(xiàng)展開式通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，則x的系數(shù)為SKIPIF1<0，故答案為：60.17．已知二項(xiàng)式SKIPIF1<0的展開式中僅有第4項(xiàng)的二項(xiàng)式系數(shù)最大，則SKIPIF1<0.【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)展開式的二項(xiàng)式系數(shù)的性質(zhì)，即可求解.【詳解】因?yàn)槎?xiàng)式SKIPIF1<0的展開式中僅有第4項(xiàng)的二項(xiàng)式系數(shù)最大，根據(jù)二項(xiàng)展開式的性質(zhì)，可得中間項(xiàng)的二項(xiàng)式系數(shù)最大，所以展開式一共有7項(xiàng)，所以SKIPIF1<0為偶數(shù)且SKIPIF1<0，可得SKIPIF1<0.故答案為：SKIPIF1<0.18．已知SKIPIF1<0的展開式中第7項(xiàng)和第8項(xiàng)的二項(xiàng)式系數(shù)相等，求展開式中系數(shù)最大的項(xiàng)及二項(xiàng)式系數(shù)最大的項(xiàng)．【答案】答案見解析【分析】利用二項(xiàng)式系數(shù)相等可得SKIPIF1<0的值，再利用二項(xiàng)式系數(shù)的性質(zhì)可得二項(xiàng)式系數(shù)最大的項(xiàng)，利用不等式法可求得系數(shù)最大的項(xiàng)，從而得解.【詳解】因?yàn)镾KI