第2课时指数幂及其运算课时过关·能力提升基础巩固1.下列各式正确的是()A.a-35=13√a5B.3√x2=x32C.a12·a14·a-18=a12×14×(-18)D.2x-13(12x13-2x-23)=1−4x答案:D2.将❑√2❑√2❑√2化为分数指数幂为()A.232B.234C.274D.278解析:❑√2❑√2❑√2=❑√2❑√2×212=❑√2×(232)12=❑√2×234=¿答案:D3.(112)0−(1−0.5−2)÷(278)23的值为()A.73B.43C.13D.−13解析:原式=1-(1-22)÷(32)2=1−(−3)×49=73.答案:A4.化简(a23b12)(−3a12b13)÷(13a16b56)的结果是()A.6aB.-aC.-9aD.9a2解析:原式=-3a23+12b12+13÷(13a16b56)=-9a23+12-16b12+13-56=−9a.答案:C5.若(a-2)-14有意义,则实数a的取值范围是()A.a≥2B.a≤2C.a>2D.a<2解析:∵(a-2)-14=14√a-2,∴若(a-2)-14有意义,则a-2>0,即a>2.答案:C6.若3a·9b¿13,则下列等式正确的是()A.a+b=-1B.a+b=1C.a+2b=-1D.a+2b=1解析:3a·9b=3a·32b=3a+2b¿13=3−1,则a+2b=-1.答案:C7.4❑√2+1×23-2❑√2×64-23=¿¿解析:原式=(22)❑√2+1×23-2❑√2׿答案:28.化简❑√a3b23√ab2(a14b12)4a-13b13(a>0,b>0)=¿.解析:原式=(a3b2a13b23)12ab2a-13b13=a32+16-1+13b1+13-2-13=ab−1=ab.答案:ab9.已知3a2+b=1,则9a·3b❑√3a=¿¿解析:9a·3b❑√3a=32a·3b3a2=32a+b-a2=33a2+b.∵3a2+b=1,∴9a·3b❑√3a=3.答案:310.已知a2+a-2=3,则a+a-1=.解析:∵a2+a-2=(a+a-1)2-2,∴(a+a-1)2-2=3,∴(a+a-1)2=5,∴a+a-1=±❑√5.答案:±❑√511.计算:(1¿(214)0.5−0.752+6−2×(827)-23;(2)(0.25)-12−[-2×(20162017)0]2׿(3)(7+4❑√3)12−8118+3235−2×(18)-23+3√2×(4-13)−1.解:(1¿(214)0.5−0.752+6−2×(827)-23¿[(32)2]12−(34)2+136×[(23)3]-23¿32−(34)2+136×(23)-2¿32−916+136×94=1.(2)(0.25)-12−[-2×(20162017)0]2׿=2-4×14+10(2+❑√3)−10❑√3=21.(3)(7+4❑√3)12−8118+3235−2×(18)-23+3√2×(4-13)−1=[(2+❑√3¿2]12−¿=2+❑√3−❑√3+8−8+2=4.12.已知a,b是方程x2-6x+4=0的两个根,且a>b>0,求❑√a-❑√b❑√a+❑√b的值.分析:将要求的式子平方,化成关于a+b,ab的式子,最后求得结果.解:∵a,b是方程x2-6x+4=0的两个根,∴{a+b=6,ab=4.∵a>b>0,∴❑√a>❑√b>0.∴❑√a-❑√b❑√a+❑√b>0.又(❑√a-❑√b❑√a+❑√b)2=a+b-2❑√aba+b+2❑√ab=6-2❑√46+2❑√4=15,∴❑√a-❑√b❑√a+❑√b=❑√15=❑√55.能力提升1.当a>0时,下列式子正确的是()A.a23+a-23=0B.a32·a23=aC.a23÷a13=a2D.(a-12)2=1a解析:利用分数指数幂的性质易知D正确.答案:D2.设a12−a-12=m,则a2+1a等于()A.m2+2B.2-m2C.m2-2D.m2解析:由a12−a-12=m,两边平方得a+a-1=m2+2.故a2+1a=a+a−1=m2+2.答案:A3.若a>0,将a2❑√a·3√a2表示成分数指数幂,其结果是()A.a12B.a56C.a76D.a32解析:a2❑√a·3√a2=a2❑√a·a23=a2❑√a53=a2a53×12=a2·a-56=a2-56=a76.答案:C4.★计算(2n+1)2·(12)2n+14n·8-2¿∈N*)的结果为()A.164B.22n+5C.2n2-2n+6D.(12)2n-7解析:原式=22n+2·2-2n-1(22)n·(23)-2=2122n-6=27−2n=(12)2n-7.答案:D5.如果a=3,b=384,那么a[(ba)17]n-3=¿¿解析:a[(ba)17]n-3=3[(3843)17]n-3=3[(128)17]n-3=3×2n−3.答案:3×2n-36.已知a<0,则化简3√6√(-a)9的结果为¿解析:∵a<0,∴3√6√(-a)9=3√(-a)96=¿答案:❑√-a7.已知10α=2,10β=3,则1003α2-β=¿.解析:1003α2-β=103α102β=(10α)3(10β)2=2332=89.答案:898.已知2x+2-x=a(a≥2),求4x+4-x的值.解:∵2x+2-x=a(a≥2),∴22x+2+2-2x=a2.∴4x+4-x=22x+2-2x=a2-2.9.★(1)计算:0.0001-14+2723−(4964)-12+(19)-1.5;(2)化简:m12-n12m12+n12+m12+n12m12-n12¿≠n).解:(1)原式=(0.14)-14+¿¿0.1−1+32−(78)-1+(13)-3=10+9−87+27=3147.(2)原式=(m12-n12)2+(m12+n12)2(m12+n12)(m12-n12)=2(m+n)m-n.
