人教A版高中數(shù)學(xué)必修第二冊(cè)同步講義第19講 復(fù)數(shù)的乘、除運(yùn)算（含解析）

第19課復(fù)數(shù)的乘、除運(yùn)算目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航課程標(biāo)準(zhǔn)課標(biāo)解讀1.掌握復(fù)數(shù)代數(shù)形式的乘法和除法運(yùn)算.2.理解復(fù)數(shù)乘法的交換律、結(jié)合律和乘法對(duì)加法的分配律．1.在熟悉課本能容的基礎(chǔ)上，掌握復(fù)數(shù)代數(shù)形式的乘法和除法運(yùn)算.2.在學(xué)習(xí)中逐步加強(qiáng)理解復(fù)數(shù)乘法的交換律、結(jié)合律和乘法對(duì)加法的分配律．知識(shí)精講知識(shí)精講知識(shí)點(diǎn)01復(fù)數(shù)乘法的運(yùn)算法則和運(yùn)算律1．復(fù)數(shù)的乘法法則設(shè)z1＝a＋bi，z2＝c＋di(a，b，c，d∈R)是任意兩個(gè)復(fù)數(shù)，則z1·z2＝(a＋bi)(c＋di)＝(ac－bd)＋(ad＋bc)i.2．復(fù)數(shù)乘法的運(yùn)算律對(duì)任意復(fù)數(shù)z1，z2，z3∈C，有交換律z1z2＝z2z1結(jié)合律(z1z2)z3＝z1(z2z3)乘法對(duì)加法的分配律z1(z2＋z3)＝z1z2＋z1z3【即學(xué)即練1】計(jì)算下列各題．(1)(1－i)(1＋i)＋(2＋i)2；(2)(2－i)(－1＋5i)(3－4i)＋2i.解(1)(1－i)(1＋i)＋(2＋i)2＝1－i2＋4＋4i＋i2＝5＋4i.(2)(2－i)(－1＋5i)(3－4i)＋2i＝(－2＋10i＋i－5i2)(3－4i)＋2i＝(3＋11i)(3－4i)＋2i＝(9－12i＋33i－44i2)＋2i＝53＋21i＋2i＝53＋23i.反思感悟(1)兩個(gè)復(fù)數(shù)代數(shù)形式的乘法運(yùn)算的一般步驟①首先按多項(xiàng)式的乘法展開．②再將i2換成－1.③然后再進(jìn)行復(fù)數(shù)的加、減運(yùn)算．(2)常用公式①(a＋bi)2＝a2－b2＋2abi(a，b∈R)．②(a＋bi)(a－bi)＝a2＋b2(a，b∈R)．③(1±i)2＝±2i.知識(shí)點(diǎn)02復(fù)數(shù)的除法法則設(shè)z1＝a＋bi，z2＝c＋di(a，b，c，d∈R，且c＋di≠0)是任意兩個(gè)復(fù)數(shù)，則eq\f(z1,z2)＝eq\f(a＋bi,c＋di)＝eq\f(a＋bic－di,c＋dic－di)＝eq\f(ac＋bd,c2＋d2)＋eq\f(bc－ad,c2＋d2)i.復(fù)數(shù)的除法的實(shí)質(zhì)是分母實(shí)數(shù)化．若分母為a＋bi型，則分子、分母同乘a－bi；若分母為a－bi型，則分子、分母同乘a＋bi，即分子分母同乘以分母的共軛復(fù)數(shù)．【即學(xué)即練2】計(jì)算：eq\f(1＋i4＋3i,2－i1－i)＝________.答案－2＋i解析方法一eq\f(1＋i4＋3i,2－i1－i)＝eq\f(1＋7i,1－3i)＝eq\f(1＋7i1＋3i,10)＝－2＋i.方法二eq\f(1＋i4＋3i,2－i1－i)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1＋i,1－i)))eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(4＋3i,2－i)))＝eq\f(i4＋3i2＋i,5)＝eq\f(－3＋4i2＋i,5)＝eq\f(－10＋5i,5)＝－2＋i.反思感悟復(fù)數(shù)的除法運(yùn)算法則的應(yīng)用復(fù)數(shù)的除法法則在實(shí)際操作中不方便使用，一般將除法寫成分式形式，采用分母“實(shí)數(shù)化”的方法，即將分子、分母同乘分母的共軛復(fù)數(shù)，使分母成為實(shí)數(shù)，再計(jì)算．能力拓展能力拓展考法01復(fù)數(shù)代數(shù)形式的乘法運(yùn)算【典例1】計(jì)算：(1－i)2－(2－3i)(2＋3i)等于()A．2i－13 B．13＋2iC．13－2i D．－13－2i答案D解析(1－i)2－(2－3i)(2＋3i)＝1－2i＋i2－(4－9i2)＝－13－2i.【變式訓(xùn)練】若復(fù)數(shù)(1－i)(a＋i)在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第二象限，則實(shí)數(shù)a的取值范圍是()A．(－∞，1) B．(－∞，－1)C．(1，＋∞) D．(－1，＋∞)答案B解析因?yàn)閦＝(1－i)(a＋i)＝a＋1＋(1－a)i，所以它在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為(a＋1,1－a)，又此點(diǎn)在第二象限，所以eq\b\lc\{\rc\(\a\vs4\al\co1(a＋1<0，,1－a>0，))解得a<－1.考法02復(fù)數(shù)代數(shù)形式的除法運(yùn)算【典例2】設(shè)復(fù)數(shù)z滿足eq\f(1＋z,1－z)＝i，則|z|等于()A．1B.eq\r(2)C.eq\r(3)D．2答案A解析由eq\f(1＋z,1－z)＝i得1＋z＝i(1－z)，即z＝eq\f(－1＋i,1＋i)＝eq\f(－1＋i1－i,1＋i1－i)＝eq\f(－1－i2,2)＝i，|z|＝1.【變式訓(xùn)練】)若復(fù)數(shù)z滿足z(2－i)＝11＋7i(i為虛數(shù)單位)，則z為()A．3＋5i B．3－5iC．－3＋5i D．－3－5i答案A解析∵z(2－i)＝11＋7i，∴z＝eq\f(11＋7i,2－i)＝eq\f(11＋7i2＋i,2－i2＋i)＝eq\f(15＋25i,5)＝3＋5i.考法03在復(fù)數(shù)范圍內(nèi)解方程【典例3】在復(fù)數(shù)范圍內(nèi)解方程x2＋6x＋10＝0.解方法一因?yàn)閤2＋6x＋10＝x2＋6x＋9＋1＝(x＋3)2＋1＝0，所以(x＋3)2＝－1，又因?yàn)閕2＝－1，所以(x＋3)2＝i2，所以x＋3＝±i，即x＝－3±i.方法二因?yàn)棣ぃ?2－4×10×1＝－4<0，所以方程的根為x＝eq\f(－6±\r(4)i,2)＝－3±i.反思感悟在復(fù)數(shù)范圍內(nèi)，實(shí)系數(shù)一元二次方程ax2＋bx＋c＝0(a≠0)的求解方法(1)求根公式法①當(dāng)Δ≥0時(shí)，x＝eq\f(－b±\r(b2－4ac),2a).②當(dāng)Δ<0時(shí)，x＝eq\f(－b±\r(－b2－4ac)i,2a).(2)利用復(fù)數(shù)相等的定義求解設(shè)方程的根為x＝m＋ni(m，n∈R)，將此根代入方程ax2＋bx＋c＝0(a≠0)，化簡(jiǎn)后利用復(fù)數(shù)相等的定義求解．【變式訓(xùn)練】已知1＋i是方程x2＋bx＋c＝0(b，c為實(shí)數(shù))的一個(gè)根．(1)求b，c的值；(2)試判斷1－i是不是方程的根．解(1)∵1＋i是方程x2＋bx＋c＝0的根，且b，c為實(shí)數(shù)，∴(1＋i)2＋b(1＋i)＋c＝0，即b＋c＋(b＋2)i＝0，∴eq\b\lc\{\rc\(\a\vs4\al\co1(b＋c＝0，,2＋b＝0，))解得eq\b\lc\{\rc\(\a\vs4\al\co1(b＝－2，,c＝2.))(2)由(1)知方程為x2－2x＋2＝0，把1－i代入方程左邊得(1－i)2－2(1－i)＋2＝0＝右邊，即方程式成立．∴1－i是方程的根．分層提分分層提分題組A基礎(chǔ)過關(guān)練一、單選題1．已知i為虛數(shù)單位，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】A【詳解】SKIPIF1<0.故選：A2．復(fù)數(shù)SKIPIF1<0的虛部為（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)镾KIPIF1<0所以虛部為1.故選：B3．已知復(fù)數(shù)SKIPIF1<0滿足：SKIPIF1<0（i為虛數(shù)單位），則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．2【答案】C【詳解】SKIPIF1<0，故SKIPIF1<0.故選：C.4．已知復(fù)數(shù)SKIPIF1<0為SKIPIF1<0的共軛復(fù)數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0得SKIPIF1<0代入計(jì)算可得SKIPIF1<0．故選：D．5．已知SKIPIF1<0，則復(fù)數(shù)z+5的實(shí)部與虛部的和為（）A．10 B．SKIPIF1<0 C．0 D．SKIPIF1<0【答案】A【詳解】由SKIPIF1<0可得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的實(shí)部與虛部的和為SKIPIF1<0，故選：A6．若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，故選：A7．已知復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.故選：A8．復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)镾KIPIF1<0，故對(duì)應(yīng)的點(diǎn)的坐標(biāo)為SKIPIF1<0.故選：B.9．已知復(fù)數(shù)SKIPIF1<0，則z的虛部是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】∵SKIPIF1<0，∴z的虛部為SKIPIF1<0.故選：B.10．已知SKIPIF1<0，則復(fù)數(shù)z的虛部為（

）A．1 B．-i C．-1 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，即SKIPIF1<0，故虛部為-1故選：C二、填空題11．設(shè)復(fù)數(shù)z滿足SKIPIF1<0（其中SKIPIF1<0是虛數(shù)單位），則SKIPIF1<0___________.【答案】SKIPIF1<0【詳解】由已知條件得SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.12．如果SKIPIF1<0與SKIPIF1<0（i為虛數(shù)單位）是共軛復(fù)數(shù)，則實(shí)數(shù)x＝_____，y＝________.【答案】

SKIPIF1<0##0.25

SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0與SKIPIF1<0（i為虛數(shù)單位）是共軛復(fù)數(shù)，所以SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.題組B能力提升練一、單選題1．已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．2 D．4【答案】A【詳解】SKIPIF1<0，SKIPIF1<0故選：A2．已知i為虛數(shù)單位，復(fù)數(shù)z滿足SKIPIF1<0，則復(fù)數(shù)z在復(fù)平面上的點(diǎn)位于（

）A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】D【詳解】由SKIPIF1<0，所以SKIPIF1<0，所以z在復(fù)平面上的點(diǎn)位于第四象限.故選：D．3．已知SKIPIF1<0為虛數(shù)單位)，則SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在（

）A．第一象限 B．第二象限C．第三象限 D．第四象限【答案】B【詳解】由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，故復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第二象限，故選：B4．已知復(fù)數(shù)SKIPIF1<0(i為虛數(shù)單位)，則復(fù)數(shù)z在復(fù)平面上對(duì)應(yīng)的點(diǎn)位于（

）A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】D【詳解】解：∵SKIPIF1<0，∴復(fù)數(shù)SKIPIF1<0又SKIPIF1<0則對(duì)應(yīng)點(diǎn)坐標(biāo)為SKIPIF1<0位于第四象限.故選：D5．設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由題意可得SKIPIF1<0.故選：B6．已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，則SKIPIF1<0故選：C.7．若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0.故選：D.8．已知SKIPIF1<0，其中SKIPIF1<0為虛數(shù)單位，則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)的坐標(biāo)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)閺?fù)數(shù)SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)的坐標(biāo)是SKIPIF1<0，故選：SKIPIF1<0.9．已知SKIPIF1<0，則SKIPIF1<0（

）A．1 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：SKIPIF1<0故選：A10．已知SKIPIF1<0（其中SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故選：A.11．若復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則下列結(jié)論正確的有（

）A．SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0的虛部為SKIPIF1<0 D．在復(fù)平面內(nèi)SKIPIF1<0是第三象限的點(diǎn)【答案】B【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0；SKIPIF1<0；SKIPIF1<0的虛部為SKIPIF1<0；在復(fù)平面內(nèi)SKIPIF1<0對(duì)應(yīng)點(diǎn)的坐標(biāo)為SKIPIF1<0，它在第四象限，故選：B二、填空題12．已知復(fù)數(shù)SKIPIF1<0，則SKIPIF1<0_______【答案】SKIPIF1<0【詳解】SKIPIF1<0故答案為：SKIPIF1<013．已知復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0___________.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.14．已知SKIPIF1<0是實(shí)數(shù)，SKIPIF1<0是虛數(shù)單位，若復(fù)數(shù)SKIPIF1<0的實(shí)部和虛部互為相反數(shù)，則SKIPIF1<0___________.【答案】SKIPIF1<0【詳解】由題意SKIPIF1<0，因?yàn)閷?shí)部和虛部互為相反數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0，則SKIPIF1<0，故答案為：SKIPIF1<015．若SKIPIF1<0是虛數(shù)單位，則復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)的坐標(biāo)為________．【答案】SKIPIF1<0【詳解】SKIPIF1<0,則復(fù)數(shù)SKIPIF1<0對(duì)應(yīng)的點(diǎn)的坐標(biāo)為SKIPIF1<0，故答案為：SKIPIF1<0．16．若復(fù)數(shù)z滿足SKIPIF1<0（i為虛數(shù)單位），則SKIPIF1<0__________．【答案】SKIPIF1<0【詳解】∵SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<0．題組C培優(yōu)拔尖練一、單選題1．已知復(fù)數(shù)z滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，SKIPIF1<0故選：B2．已知復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)閺?fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A.3．復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)向量SKIPIF1<0的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，所以SKIPIF1<0，故選：B.4．SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0則SKIPIF1<0故選:A5．已知復(fù)數(shù)滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)閺?fù)數(shù)滿足SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0SKIPIF1<0.故選：B6．已知復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則SKIPIF1<0的共軛復(fù)數(shù)的模是（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的共軛復(fù)數(shù)為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的共軛復(fù)數(shù)的模是SKIPIF1<0.故選：C7．若SKIPIF1<0，則SKIPIF1<0的實(shí)部為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的實(shí)部為SKIPIF1<0.故選：C.二、多選題8．已知非零復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)分別為SKIPIF1<0為坐標(biāo)原點(diǎn)，則（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．若SKIPIF1<0，則存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<0【答案】AC【解答】對(duì)A，SKIPIF1<0即SKIPIF1<0，兩邊平方可得SKIPIF1<0，A對(duì)；對(duì)SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0，B錯(cuò)；對(duì)SKIPIF1<0，SKIPIF1<0即SKIPIF1<0，兩邊平方可得SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，因此存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，C對(duì)；對(duì)SKIPIF1<0，取SKIPIF1<0，但SKIPIF1<0，D錯(cuò).故選：AC9．下列說法中正確的有（

）A．已知復(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)在第四象限；B．已知復(fù)數(shù)SKIPIF1<0（SKIPIF1<0為虛數(shù)單位），則復(fù)數(shù)SKIPIF1<0在復(fù)平面內(nèi)所對(duì)應(yīng)的點(diǎn)在第三象限；C．在SKIPIF1<0中，若SKIPIF1<0，則SKIPIF1<0為等腰或直角三角形；D．在SKIPIF1<0中，若SKIPIF1<0，則SKIPIF1<0為等腰三角形.【答案】ABD【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，其所對(duì)應(yīng)的點(diǎn)的坐標(biāo)是SKIPIF1<0，在第四象限，故A正確；SKIPIF1<0，所對(duì)應(yīng)的點(diǎn)的坐標(biāo)是SKIPIF1<0，在第三象限，故B正確；因?yàn)镾KIPIF1<0，結(jié)合正弦定理可得SKIPIF1<0，因此SKIPIF1<0為等腰三角形，故C錯(cuò)誤；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0