1、2指数幂的运算性质课后训练巩固提升一、A组1.(-1)-22等于()A.-1B.1C.2D.-2解析:(-1)-22=1(-1)22=12=1.答案:B2.设a0,将aa3a2表示成分数指数幂的形式,其结果是()A.a16B.a56C.a76D.a32解析:原式=aa12a223=aa12+13=a1-56=a16.答案:A3.(36a9)4(63a9)4(a0)的化简结果是()A.a16B.a8C.a4D.a2解析:原式=(36a963a9)4=(3a966a93)4=(3a326a3)4=(a12a12)4=a4.答案:C4.a3a5a4(a0)的值是()A.1B.aC.a15D.a171
2、0解析:原式=a3a12a45=a3-12-45=a1710.答案:D5.若10x=3,10y=4,则102x-y=.解析:因为10x=3,所以(10x)2=9,即102x=9,所以102x10y=94,即102x-y=94.答案:946.计算(323)6+(22)43-41649-12-4280.25-(2 005)0=.解析:原式=(213312)6+(212214)43-474-214234-1=2233+2-7-2-1=100.答案:1007.化简(2n+1)2122n+14n8-2(nN+)的结果为.解析:原式=22n+22-2n-122n2-6=22n+2-2n-1-2n+6=27
3、-2n.答案:27-2n8.计算下列各式:(1)2790.5+(0.1)-2+21027-23+30+3748;(2)525535125;(3)0.025 6-14-78-2.60+(34)34(22)53-160.75.解:(1)原式=2590.5+110-2+6427-23+3+3748=53+100+916+3+3748=80+27+3748+103=14448+103=3+103=106.(2)原式=525355-125-32=52+35-12-32=535.(3)原式=2.5-1+2233423253-23=1.5+212+52-23=1.5.二、B组1.下列各式成立的是()A.mn
4、7=m7n17(m,n0)B.12(-3)4=3-3C.4x3+y3=(x+y)34(x,y0)D.39=33解析:mn7=m7n7=m7n-7m7n17;12(-3)4=1234=333-3;4x3+y3=(x3+y3)14(x+y)34;39=(32)1312=(32)16=3216=313=33.答案:D2.若a2x=2-1,则a3x+a-3xax+a-x=()A.22-1B.2-22C.22+1D.2+1解析:a3x+a-3xax+a-x=(ax+a-x)(a2x-1+a-2x)ax+a-x=a2x+1a2x-1=2-1+12-1-1=22-1.答案:A3.若x=1+2b,y=1+2-
5、b,则y等于()A.x+1x-1B.x-1xC.x-1x+1D.xx-1解析:x=1+2b,2b=x-1.又y=1+2-b=1+12b=2b+12b=x-1+1x-1=xx-1.答案:D4.已知a0,b0,则a3b3ab=()A.a16b76B.a76b16C.a13b16D.a12b16解析:a3b3ab=a32b12a13b13=a76b16.答案:B5.已知实数x满足5x-1103x=8x,则x=.解析:根据题意,5x-1103x=8x,即5x-1(25)3x=23x,则有54x-1=1,即4x-1=0,解得x=14.答案:146.若a1,b0,且ab+a-b=22,求ab-a-b的值.解:因为ab+a-b=(ab2+a-b2)2-2,所以(ab2+a-b2)2=ab+a-b+2=2(2+1).又ab2+a-b20,所以ab2+a-b2=2(2+1);由于a1,b0,则ab2a-b2,即ab2-a-b20,同理可得ab2-a-b2=2(2-1),得ab-a-b=2.7.已知a0,b0,且ab=ba,求证:abab=aa-bb.证明:由ab=ba知b=aba,则abab=aabbab=aab(aba)ab=aab-1=aa-bb=右边.所以等式成立.
