高一數(shù)學(xué)指數(shù)與指數(shù)冪的運(yùn)算訓(xùn)練題

字號(hào):

為大家整理的高一數(shù)學(xué)指數(shù)與指數(shù)冪的運(yùn)算訓(xùn)練題文章,供大家學(xué)習(xí)參考!更多最新信息請(qǐng)點(diǎn)擊高一考試網(wǎng)
    1.將532寫(xiě)為根式,則正確的是(  )
    A.352        B.35
    C.532 D.53
    解析:選D.532=53.
    2.根式 1a1a(式中a>0)的分?jǐn)?shù)指數(shù)冪形式為(  )
    A.a(chǎn)-43 B.a(chǎn)43
    C.a(chǎn)-34 D.a(chǎn)34
    解析:選C.1a1a= a-1•a-112= a-32=(a-32)12=a-34.
    3.a-b2+5a-b5的值是(  )
    A.0 B.2(a-b)
    C.0或2(a-b) D.a(chǎn)-b
    解析:選C.當(dāng)a-b≥0時(shí),
    原式=a-b+a-b=2(a-b);
    當(dāng)a-b<0時(shí),原式=b-a+a-b=0.
    4.計(jì)算:(π)0+2-2×(214)12=________.
    解析:(π)0+2-2×(214)12=1+122×(94)12=1+14×32=118.
    答案:118
    1.下列各式正確的是(  )
    A.-32=-3 B.4a4=a
    C.22=2 D.a(chǎn)0=1
    解析:選C.根據(jù)根式的性質(zhì)可知C正確.
    4a4=|a|,a0=1條件為a≠0,故A,B,D錯(cuò).
    2.若(x-5)0有意義,則x的取值范圍是(  )
    A.x>5 B.x=5
    C.x<5 D.x≠5
    解析:選D.∵(x-5)0有意義,
    ∴x-5≠0,即x≠5.
    3.若xy≠0,那么等式 4x2y3=-2xyy成立的條件是(  )
    A.x>0,y>0 B.x>0,y<0
    C.x<0,y>0 D.x<0,y<0
    解析:選C.由y可知y>0,又∵x2=|x|,
    ∴當(dāng)x<0時(shí),x2=-x.
    4.計(jì)算2n+12•122n+14n•8-2(n∈N*)的結(jié)果為(  )
    A.164 B.22n+5
    C.2n2-2n+6 D.(12)2n-7
    解析:選D.2n+12•122n+14n•8-2=22n+2•2-2n-122n•23-2=2122n-6=27-2n=(12)2n-7.
    5.化簡(jiǎn) 23-610-43+22得(  )
    A.3+2 B.2+3
    C.1+22 D.1+23
    解析:選A.原式= 23-610-42+1
    = 23-622-42+22= 23-62-2
    = 9+62+2=3+2.X k b 1 . c o m
    6.設(shè)a12-a-12=m,則a2+1a=(  )
    A.m2-2 B.2-m2
    C.m2+2 D.m2
    解析:選C.將a12-a-12=m平方得(a12-a-12)2=m2,即a-2+a-1=m2,所以a+a-1=m2+2,即a+1a=m2+2⇒a2+1a=m2+2.
    7.根式a-a化成分?jǐn)?shù)指數(shù)冪是________.
    解析:∵-a≥0,∴a≤0,
    ∴a-a=--a2-a=--a3=-(-a)32.
    答案:-(-a)32
    8.化簡(jiǎn)11+62+11-62=________.
    解析: 11+62+11-62=3+22+3-22=3+2+(3-2)=6.
    答案:6
    9.化簡(jiǎn)(3+2)2010•(3-2)2011=________.
    解析:(3+2)2010•(3-2)2011
    =[(3+2)(3-2)]2010•(3-2)
    =12010•(3-2)= 3-2.
    答案:3-2
    10.化簡(jiǎn)求值:
    (1)0.064-13-(-18)0+1634+0.2512;
    (2)a-1+b-1ab-1(a,b≠0).
    解:(1)原式=(0.43)-13-1+(24)34+(0.52)12
    =0.4-1-1+8+12
    =52+7+12=10.
    (2)原式=1a+1b1ab=a+bab1ab=a+b.
    11.已知x+y=12,xy=9,且x    解:x12-y12x12+y12=x+y-2xy12x-y.
    ∵x+y=12,xy=9,
    則有(x-y)2=(x+y)2-4xy=108.
    又x    代入原式可得結(jié)果為-33.
    12.已知a2n=2+1,求a3n+a-3nan+a-n的值.
    解:設(shè)an=t>0,則t2=2+1,a3n+a-3nan+a-n=t3+t-3t+t-1
    =t+t-1t2-1+t-2t+t-1=t2-1+t-2
    =2+1-1+12+1=22-1.