度高一數(shù)學(xué)人教A版必修一練習(xí)第二章基本初等函數(shù)（Ⅰ）2.1.1第二課時(shí)指數(shù)冪及其運(yùn)算性質(zhì)

第二課時(shí)指數(shù)冪及其運(yùn)算性質(zhì)【選題明細(xì)表】知識(shí)點(diǎn)、方法題號(hào)根式與指數(shù)冪互化1,4,5利用指數(shù)冪的運(yùn)算性質(zhì)化簡(jiǎn)求值2,3,6,8,9,10,12,13,14,15附加條件的冪的求值問題7,111.將a·3a(A)a16 (B)a56 (C)a解析:a·3a=a12·a132.下列運(yùn)算中,正確的是(A)(A)x3·x2=x5 (B)x+x2=x3(C)2x3÷x2=x (D)(x2)3=解析:對(duì)于A,根據(jù)同底數(shù)的運(yùn)算法則可得,x3·x2=x5,故正確;對(duì)于B,不是同類項(xiàng),不能合并,故錯(cuò)誤;C,2x3÷x2=2x3D,(x2)3=x3.(112)0(10.52)÷(278)(A)13 (B)13 (C)43 解析:原式=1(14)÷3(278)

24.下列各式中成立的一項(xiàng)是(D)(A)(nm)7=n7m17 (B)(C)4x3+y3=(x+y解析:A中(nm)7=n7m7,故A錯(cuò);B中的12(-3)4=3412=313=335.設(shè)a>0,將a2(A)a12 (B)a56 (C)a解析:由題意a2a·3a6.[810.25+(338)

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12+解析:[810.25+(338)

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12+12lg4lg15=[(34)0.25+(32)

答案:27.若a+b=3,則代數(shù)式a3+b3+9ab的值為.

解析:因?yàn)閍+b=3,所以代數(shù)式a3+b3+9ab=(a+b)(a2+b2ab)+9ab=3(a2答案:278.a3b23解析:原式=(a3b2=ab1=ab答案:a9.計(jì)算:求(214)

12(9.6)0(338)解:原式=321(32)

=1249=1210.(1)計(jì)算:92×32×6(2)已知x+x1=3(x>0),求x32+解:(1)原式=321(2)因?yàn)閤+x1=3,所以x2+x2=7,所以(x32+x=x3+x3+2=(x+x1)(x2+x21)+2=3×6+2=20,所以x32+x-11.若x+x1=3,那么x2x2的值為(A)(A)±35 (B)5 (C)35 (D)13解析:因?yàn)閤+x1=3,所以(x+x1)2=x2+x2+2=9,所以x2+x2=7.所以(xx1)2=x2+x22=5,所以xx1=±5.當(dāng)xx1=5時(shí),x2x2=(x+x1)(xx1)=35,當(dāng)xx1=5時(shí),x2x2=(x+x1)(xx1)=35.故選A.12.設(shè)a12a-12=m,則解析:將a12a-12=m平方得(a1即a2+a1=m2,所以a+a1=m2+2,即a+1a=m2+2?a2+1答案:m2+213.計(jì)算:0.064-13(18)0+1634解析:原式=0.43×(-=2.51+8+0.5=10.答案:1014.計(jì)算下列各式的值:(1)1.5-13×(76)0+80.25×42+(32(2)3a72a-解:(1)原式=(23)

13×1+(23)14×214+(213×3(2)原式=3a72a-32÷a-83a153÷3a-32a-12=315.(1)化簡(jiǎn):3xy2·xy-1(2)計(jì)算:2-12+(-4)