新高考數(shù)學(xué)三輪沖刺天津卷押題練習(xí)第10~11題（教師版）

復(fù)數(shù)計(jì)算、二項(xiàng)式定理考點(diǎn)2年考題考情分析復(fù)數(shù)計(jì)算2023年天津卷第10題2022年天津卷第10題高考對復(fù)數(shù)知識(shí)的考查要求較低，一般難度不大，要求考生熟練復(fù)數(shù)基礎(chǔ)知識(shí)點(diǎn)，包括復(fù)數(shù)的代數(shù)形式，復(fù)數(shù)的實(shí)部與虛部，共軛復(fù)數(shù)，復(fù)數(shù)模長，復(fù)數(shù)的幾何意義及四則運(yùn)算．可以預(yù)測2024年高考命題方向?qū)⒗^續(xù)圍繞復(fù)數(shù)的四則運(yùn)算為背景展開命題．二項(xiàng)式定理2023年天津卷第11題2022年天津卷第11題高考對二項(xiàng)式定理知識(shí)的考察要求較低，一般難度不大，要求學(xué)生掌握二項(xiàng)式定理的展開式運(yùn)算，會(huì)計(jì)算組合數(shù)以及冪的化簡運(yùn)算?？梢灶A(yù)測2024年高考命題方向?qū)⒗^續(xù)圍繞二項(xiàng)式的展開式中某一項(xiàng)的系數(shù)為背景展開命題．題型一復(fù)數(shù)運(yùn)算10．（5分）（2023?天津）已知SKIPIF1<0是虛數(shù)單位，化簡SKIPIF1<0的結(jié)果為SKIPIF1<0．【答案】SKIPIF1<0．【分析】直接利用復(fù)數(shù)代數(shù)形式的乘除運(yùn)算化簡求解即可．【解答】解：SKIPIF1<0．故答案為：SKIPIF1<0．10．（5分）（2022?天津）已知SKIPIF1<0是虛數(shù)單位，化簡SKIPIF1<0的結(jié)果為SKIPIF1<0．【答案】SKIPIF1<0．【分析】直接利用復(fù)數(shù)代數(shù)形式的乘除運(yùn)算化簡即可．【解答】解：SKIPIF1<0，故答案為：SKIPIF1<0．一、復(fù)數(shù)的概念=1\*GB3①復(fù)數(shù)的概念：形如a＋bi(a，b∈R)的數(shù)叫做復(fù)數(shù)，a，b分別是它的實(shí)部和虛部，SKIPIF1<0叫虛數(shù)單位，滿足SKIPIF1<0（1）當(dāng)且僅當(dāng)b＝0時(shí)，a＋bi為實(shí)數(shù)；（2）當(dāng)b≠0時(shí)，a＋bi為虛數(shù)；（3）當(dāng)a＝0且b≠0時(shí)，a＋bi為純虛數(shù)．其中，兩個(gè)實(shí)部相等，虛部互為相反數(shù)的復(fù)數(shù)互為共軛復(fù)數(shù)．=2\*GB3②兩個(gè)復(fù)數(shù)SKIPIF1<0相等SKIPIF1<0（兩復(fù)數(shù)對應(yīng)同一點(diǎn)）=3\*GB3③復(fù)數(shù)的模：復(fù)數(shù)SKIPIF1<0的模，其計(jì)算公式SKIPIF1<0二、復(fù)數(shù)的加、減、乘、除的運(yùn)算法則1、復(fù)數(shù)運(yùn)算（1）SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0其中SKIPIF1<0，叫z的模；SKIPIF1<0是SKIPIF1<0的共軛復(fù)數(shù)SKIPIF1<0．（3）SKIPIF1<0．實(shí)數(shù)的全部運(yùn)算律（加法和乘法的交換律、結(jié)合律、分配律及整數(shù)指數(shù)冪運(yùn)算法則）都適用于復(fù)數(shù)．1．SKIPIF1<0是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：由題意，SKIPIF1<0．故答案為：SKIPIF1<0．2．SKIPIF1<0是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：SKIPIF1<0．故答案為：SKIPIF1<0．3．已知SKIPIF1<0是純虛數(shù)（其中SKIPIF1<0，SKIPIF1<0是虛數(shù)單位），則SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：由題意SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0．4．SKIPIF1<0是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0的虛部為SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：SKIPIF1<0，其虛部為SKIPIF1<0．故答案為：SKIPIF1<0．5．SKIPIF1<0為虛數(shù)單位，復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0．故答案為：SKIPIF1<0．6．SKIPIF1<0是虛數(shù)單位，復(fù)數(shù)SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0【解答】解：SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0．7．SKIPIF1<0是復(fù)數(shù)單位，化簡SKIPIF1<0的結(jié)果為SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：SKIPIF1<0．故答案為：SKIPIF1<0．8．已知SKIPIF1<0是虛數(shù)單位，化簡SKIPIF1<0的結(jié)果為SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：SKIPIF1<0．故答案為：SKIPIF1<0．9．已知復(fù)數(shù)SKIPIF1<0（其中SKIPIF1<0為虛數(shù)單位），則SKIPIF1<05．【解答】解：SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0．故答案為：5．10．設(shè)SKIPIF1<0為虛數(shù)單位，若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0．則SKIPIF1<0SKIPIF1<0．【解答】解：SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0．故答案為：SKIPIF1<0．11．若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0是虛數(shù)單位），則SKIPIF1<0的虛部為2．【答案】2．【解答】解：SKIPIF1<0，則SKIPIF1<0的虛部為2．故答案為：2．12．若復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：因?yàn)閺?fù)數(shù)SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0．故答案為：SKIPIF1<0．13．已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0．（其中SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：由SKIPIF1<0，得SKIPIF1<0，則復(fù)數(shù)SKIPIF1<0的虛部為SKIPIF1<0．故答案為：SKIPIF1<0．14．復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【解答】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，故答案為：SKIPIF1<015．若SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【解答】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0．故答案為：SKIPIF1<0．題型二二項(xiàng)式定理11．（5分）（2023?天津）在SKIPIF1<0的展開式中，SKIPIF1<0項(xiàng)的系數(shù)為60．【答案】60．【分析】根據(jù)二項(xiàng)展開式的通項(xiàng)公式求解．【解答】解：二項(xiàng)式SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0得，SKIPIF1<0，SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0．故答案為：60．11．（5分）（2022?天津）SKIPIF1<0的展開式中的常數(shù)項(xiàng)為15．【分析】先寫出二項(xiàng)式的展開式的通項(xiàng)，整理出最簡形式，要求展開式的常數(shù)項(xiàng)，只要使得變量的指數(shù)等于0，求出SKIPIF1<0的值，代入系數(shù)求出結(jié)果．【解答】解：SKIPIF1<0SKIPIF1<0的展開式的通項(xiàng)是SKIPIF1<0要求展開式中的常數(shù)項(xiàng)只要使得SKIPIF1<0，即SKIPIF1<0SKIPIF1<0常數(shù)項(xiàng)是SKIPIF1<0，故答案為：151．二項(xiàng)式定理二項(xiàng)式定理(a＋b)n＝Ceq\o\al(0,n)an＋Ceq\o\al(1,n)an－1b1＋…＋Ceq\o\al(k,n)an－kbk＋…＋Ceq\o\al(n,n)bn(n∈N*)二項(xiàng)展開式的通項(xiàng)Tk＋1＝Ceq\o\al(k,n)an－kbk，它表示展開式的第k＋1項(xiàng)二項(xiàng)式系數(shù)Ceq\o\al(k,n)(k＝0,1，…，n)2.二項(xiàng)式系數(shù)的性質(zhì)(1)對稱性：與首末兩端“等距離”的兩個(gè)二項(xiàng)式系數(shù)相等．(2)增減性與最大值：當(dāng)n是偶數(shù)時(shí)，中間的一項(xiàng)SKIPIF1<0取得最大值；當(dāng)n是奇數(shù)時(shí)，中間的兩項(xiàng)SKIPIF1<0與SKIPIF1<0相等，且同時(shí)取得最大值．(3)各二項(xiàng)式系數(shù)的和：(a＋b)n的展開式的各二項(xiàng)式系數(shù)的和等于2n.常用結(jié)論1．兩個(gè)常用公式(1)Ceq\o\al(0,n)＋Ceq\o\al(1,n)＋Ceq\o\al(2,n)＋…＋Ceq\o\al(n,n)＝2n.(2)Ceq\o\al(0,n)＋Ceq\o\al(2,n)＋Ceq\o\al(4,n)＋…＝Ceq\o\al(1,n)＋Ceq\o\al(3,n)＋Ceq\o\al(5,n)＋…＝2n－1.2．二項(xiàng)展開式的三個(gè)重要特征(1)字母a的指數(shù)按降冪排列由n到0.(2)字母b的指數(shù)按升冪排列由0到n.(3)每一項(xiàng)字母a的指數(shù)與字母b的指數(shù)的和等于n.1．若SKIPIF1<0的展開式中常數(shù)項(xiàng)為SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：根據(jù)二項(xiàng)式的展開式：SKIPIF1<0，1，2，3，4，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，常數(shù)項(xiàng)為SKIPIF1<0．故答案為：SKIPIF1<0．2．SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)的系數(shù)是80．（用數(shù)字作答）【解答】解：在SKIPIF1<0的展開式中，通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，求得SKIPIF1<0，故SKIPIF1<0的展開式中SKIPIF1<0項(xiàng)的系數(shù)是SKIPIF1<0，故答案為80．3．在SKIPIF1<0的展開式中，常數(shù)項(xiàng)為60SKIPIF1<0（請用數(shù)字作答）【答案】60．【解答】解：二項(xiàng)式的展開式的通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0，1，2，SKIPIF1<0，6，令SKIPIF1<0，解得SKIPIF1<0，所以展開式的常數(shù)項(xiàng)為SKIPIF1<0，故答案為：60．4．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是SKIPIF1<0．【答案】SKIPIF1<0．【解答】解：二項(xiàng)式SKIPIF1<0的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0得，SKIPIF1<0，SKIPIF1<0SKIPIF1<0的系數(shù)是SKIPIF1<0，故答案為：SKIPIF1<0．5．若SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為160，則實(shí)數(shù)SKIPIF1<0的值為2．【答案】2．【解答】解：SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)為SKIPIF1<0，根據(jù)題意得SKIPIF1<0，解得SKIPIF1<0．故答案為：2．6．已知二項(xiàng)式SKIPIF1<0，則其展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為4320．【答案】4320．【解答】解：二項(xiàng)式的展開式的通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0，1，SKIPIF1<0，6，令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：4320．7．在SKIPIF1<0的二項(xiàng)展開式中，SKIPIF1<0的系數(shù)為SKIPIF1<0（請用數(shù)字作答）．【答案】SKIPIF1<0．【解答】解：二項(xiàng)式的展開式的通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0