新高考數(shù)學一輪復習6.5 二項式定理（精練）（基礎(chǔ)版）（解析版）

6.5二項式定理（精練）（基礎(chǔ)版）題組一指定項系數(shù)1．（2022·全國·高三專題練習）在SKIPIF1<0的展開式中，含SKIPIF1<0項的系數(shù)為（

）題組一指定項系數(shù)A．160 B．192 C．184 D．186【答案】B【解析】二項式SKIPIF1<0的展開式的通項SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0項的系數(shù)為192.故選：B.2．（2022·全國·高三專題練習）若SKIPIF1<0SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0SKIPIF1<0，則SKIPIF1<0.故選：A.3．（2022·陜西·武功縣普集高級中學高三階段練習（理））SKIPIF1<0的常數(shù)項的二項式系數(shù)為（

）A．375 B．-375 C．15 D．-15【答案】C【解析】由二項式SKIPIF1<0展開式的通項公式為：SKIPIF1<0；令SKIPIF1<0可得SKIPIF1<0，即展開式的中第5項是常數(shù)項.∴常數(shù)項的二項式系數(shù)為：SKIPIF1<0；故選：C.4．（2022·新疆·三模（理））若SKIPIF1<0，則SKIPIF1<0（

）A．270 B．135 C．SKIPIF1<0135 D．SKIPIF1<0270【答案】B【解析】SKIPIF1<0，以SKIPIF1<0代替SKIPIF1<0，得SKIPIF1<0，所以其通項公式為SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，故選：B5．（2022·江蘇·揚州中學高三階段練習）SKIPIF1<0的展開式中有理項的個數(shù)為_____．【答案】17【解析】通項公式SKIPIF1<0，有理項只需要保證SKIPIF1<0為整數(shù)即可，又SKIPIF1<0，故SKIPIF1<0，共17個.故答案為：17.6．（2022·上海金山·二模）SKIPIF1<0的二項展開式中SKIPIF1<0項的系數(shù)為__________.（結(jié)果用數(shù)字作答）【答案】24【解析】由題意可得SKIPIF1<0的通項公式為：SKIPIF1<0，故SKIPIF1<0項的系數(shù)為SKIPIF1<0，故答案為：247．（2022·湖北·天門市教育科學研究院模擬預(yù)測）SKIPIF1<0的展開式中的常數(shù)項為__________.【答案】SKIPIF1<0【解析】SKIPIF1<0的展開式的通項為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<0.8．（2022·全國·高三專題練習）在SKIPIF1<0的展開式中，常數(shù)項為__________.【答案】-252【解析】根據(jù)二項式定理，第r+1項為SKIPIF1<0，由于是常數(shù)，SKIPIF1<0，r=5，其常數(shù)項系數(shù)為SKIPIF1<0=-252.，故答案為：-252.9．（2022·四川·樹德中學高三開學考試（理））若二項式SKIPIF1<0SKIPIF1<0的展開式中第SKIPIF1<0項與第SKIPIF1<0項的系數(shù)相同，則其常數(shù)項是___________.【答案】SKIPIF1<0【解析】由已知條件可得SKIPIF1<0，所以，SKIPIF1<0，二項式SKIPIF1<0的展開式通項為SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，因此展開式中的常數(shù)項為SKIPIF1<0，故答案為：SKIPIF1<0.題組二題組二二項式因式相乘的系數(shù)1．（2022·內(nèi)蒙古呼倫貝爾·二模（理））SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為___________.（用數(shù)字作答）【答案】-162【解析】SKIPIF1<04屬開式的通項公式為SKIPIF1<0.當SKIPIF1<0時，xSKIPIF1<04的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0；當SKIPIF1<0時SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，故SKIPIF1<0—SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為-162.故答案為：-1622．（2022·江蘇·南京市第一中學高三開學考試）SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為___________.【答案】SKIPIF1<0【解析】SKIPIF1<0，其中SKIPIF1<0的展開式通項為SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0時，得含SKIPIF1<0的項為SKIPIF1<0；SKIPIF1<0的展開式通項為SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0時，得含SKIPIF1<0的項為SKIPIF1<0.因此，式子SKIPIF1<0的展開式中，含SKIPIF1<0的項為SKIPIF1<0，即系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<03．（2022·四川內(nèi)江·模擬預(yù)測（理））SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為______（用數(shù)字作答）.【答案】9【解析】由SKIPIF1<0的展開式通項為SKIPIF1<0，當SKIPIF1<0時SKIPIF1<0，當SKIPIF1<0時SKIPIF1<0，所以含SKIPIF1<0的項為SKIPIF1<0.故SKIPIF1<0的系數(shù)為9.故答案為：94．（2022·廣西柳州·模擬預(yù)測（理））SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為___（用數(shù)字作答）．【答案】SKIPIF1<0【解析】因為SKIPIF1<0的展開式通項為SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，在SKIPIF1<0中，令SKIPIF1<0，可得SKIPIF1<0，所以，展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.5．（2022·四川省內(nèi)江市第六中學模擬預(yù)測（理））SKIPIF1<0的展開式中的常數(shù)項為___________.（用數(shù)字作答）【答案】50【解析】因為SKIPIF1<0，考慮SKIPIF1<0中的常數(shù)項與SKIPIF1<0項.由通項公式SKIPIF1<0，即SKIPIF1<0，故當SKIPIF1<0時，SKIPIF1<0中的常數(shù)項為SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0中SKIPIF1<0的項系數(shù)為SKIPIF1<0，故SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0故答案為：SKIPIF1<06．（2022·全國·高三專題練習）SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為________________（用數(shù)字作答）．【答案】-28【解析】因為SKIPIF1<0，所以SKIPIF1<0的展開式中含SKIPIF1<0的項為SKIPIF1<0，SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為-28故答案為：-287．（2022·全國·高三專題練習）SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是12，則實數(shù)a的值為（

）A．4 B．5 C．6 D．7【答案】C【解析】利用二項式定理展開得SKIPIF1<0則SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：C.8．（2022·全國·高三專題練習）已知SKIPIF1<0的展開式中所有項的系數(shù)和為192，則展開式中的常數(shù)項為（

）A．8 B．6 C．4 D．2【答案】A【解析】令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0．在SKIPIF1<0中，SKIPIF1<0的展開式的通項SKIPIF1<0，所以SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0．故選：A9．（2022·全國·高三專題練習）SKIPIF1<0的展開式中SKIPIF1<0項的系數(shù)為（

）A．96 B．SKIPIF1<0 C．120 D．SKIPIF1<0【答案】A【解析】依題意SKIPIF1<0的展開式的通項公式為SKIPIF1<0，當SKIPIF1<0時，得SKIPIF1<0；當SKIPIF1<0時，得SKIPIF1<0，故可得展開式中含SKIPIF1<0的項為SKIPIF1<0，即展開式中SKIPIF1<0項的系數(shù)為96.故選:A10．（2022·天津·南開中學高三階段練習）SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0SKIPIF1<0，故它的展開式中含SKIPIF1<0的系數(shù)為SKIPIF1<0，故選：C．題組三題組三三項式系數(shù)1．（2022·全國·高三專題練習（理））SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是__________．【答案】120【解析】法一：在SKIPIF1<0的展開式中能產(chǎn)生SKIPIF1<0的項為SKIPIF1<0，所以在這幾項的展開式中SKIPIF1<0的系數(shù)和為SKIPIF1<0．法二：SKIPIF1<0，所以SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)就是SKIPIF1<0中SKIPIF1<0的系數(shù)，而SKIPIF1<0中SKIPIF1<0的系數(shù)為SKIPIF1<0，SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為120.故答案為：120.2．（2022·江蘇淮安·模擬預(yù)測）SKIPIF1<0的展開式的常數(shù)項是___________．【答案】70【解析】SKIPIF1<0的通項公式為SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0中的常數(shù)項為SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0中的常數(shù)項為SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0；所以SKIPIF1<0的展開式的常數(shù)項為SKIPIF1<0；故答案為：70.3．（2022·湖北·襄陽五中模擬預(yù)測）在SKIPIF1<0的展開式中，SKIPIF1<0項的系數(shù)為__．【答案】66【解析】在SKIPIF1<0的表示12個因式SKIPIF1<0的乘積，故有2個因式取SKIPIF1<0，其余的10個因式都取1，可得展開式中，含SKIPIF1<0的項，故含SKIPIF1<0項的系數(shù)為SKIPIF1<0，故答案為：664．（2022·全國·高三專題練習）SKIPIF1<0的展開式中常數(shù)項為【答案】SKIPIF1<0【解析】將原式看成6個相同的因子相乘，按x的選取個數(shù)分類，得展開式中常數(shù)項為SKIPIF1<0．5．（2022·浙江·模擬預(yù)測）在SKIPIF1<0的展開式中含SKIPIF1<0和含SKIPIF1<0的項的系數(shù)之和為【答案】SKIPIF1<0【解析】SKIPIF1<0SKIPIF1<0，則SKIPIF1<0的系數(shù)為1，SKIPIF1<0的系數(shù)為SKIPIF1<0，所以在SKIPIF1<0的展開式中含SKIPIF1<0和含SKIPIF1<0的項的系數(shù)之和為SKIPIF1<0．6．（2022·全國·高三專題練習）SKIPIF1<0的展開式中的常數(shù)項為【答案】SKIPIF1<0【解析】SKIPIF1<0，所以展開式中的常數(shù)項為SKIPIF1<07．（2022·安徽·合肥市第八中學模擬預(yù)測（理））在SKIPIF1<0的二項展開式中含SKIPIF1<0項的系數(shù)為【答案】21【解析】SKIPIF1<0的展開式的通項為SKIPIF1<0．SKIPIF1<0的展開式的通項為SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0在SKIPIF1<0的展開式中，含SKIPIF1<0項的系數(shù)為SKIPIF1<0．題組四題組四系數(shù)和1．（2022·北京工業(yè)大學附屬中學三模）若SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0【解析】在SKIPIF1<0中，令SKIPIF1<0可得：SKIPIF1<0.再令SKIPIF1<0可得：SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<0.2．（2022·遼寧沈陽·三模）若SKIPIF1<0，則SKIPIF1<0_______．【答案】243【解析】SKIPIF1<0的展開式得通項為SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0.故答案為：243.3．（2022·湖北·荊州中學模擬預(yù)測）設(shè)SKIPIF1<0．若SKIPIF1<0，則實數(shù)SKIPIF1<0________．【答案】SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0解得：SKIPIF1<0.故答案為：SKIPIF1<0.4．（2022·陜西·寶雞中學模擬預(yù)測）設(shè)SKIPIF1<0，則SKIPIF1<0