高中數(shù)學(xué)第二章基本初等函數(shù)Ⅰ2.1指數(shù)函數(shù)2.1.1指數(shù)與指數(shù)冪的運(yùn)算第1課時(shí)根式練習(xí)新人教A版必修1

PAGEPAGE12.1指數(shù)函數(shù)2.1.1指數(shù)與指數(shù)冪的運(yùn)算第1課時(shí)根式課時(shí)過關(guān)·實(shí)力提升基礎(chǔ)鞏固1.若m3=64,則m等于()A.±2 B.4 C.-2 D.2解析:∵m3=64,∴m為64的立方根,∴m=4,∴m=4=答案:D2.下列各式正確的是()A.3a3=a B.(47)C.(5a)5=|a| D.6答案:A3.化簡:1π+πA.0 B.1π+π C.1π-π D.π解析:原式=1=1=1π∵1π<π,∴原式=π-1答案:D4.3(-4)A.-8 B.8 C.0 D.1解析:3(-4)3+4(-答案:C5.(32-5)3-(45-A.25 B.4 C.4-25 D.25-4解析:(32-5)3-(45-2)4=2-5-(5-2)答案:C6.已知a<0,b∈R,則4a4-3解析:∵a<0,∴4a4又3b3∴原式=-a-b.答案:-a-b7.(6π-2)6+(54-π解析:原式=π-2+4-π=2.答案:28.若x4=81,則x=.

解析:∵x4=81,∴x為81的四次方根,∴x=±481=±3答案:±39.若12≤x≤2,則4x2-4x+1解析:∵12≤x≤2,∴2x-1≥0,x-∴4x2-=(2x-1)2+2|x-2|=2x-1-答案:310.已知(36+m)3+4(5-m)4解析:(36+m)3+4(5-m)4=6+m+|5-m|.當(dāng)m>5時(shí),|5-m|=m-5,此時(shí),6+m+|5-m|=6+m+m-5=2m+1>11;當(dāng)m≤5時(shí),|5-m|=5-m,此時(shí),6+m+|5-m|=6+m+5-m=11.所以答案:(-∞,5]11.求614解:原式=254-3278+3實(shí)力提升1.已知xy≠0,且4x2y2=-2A.xy<0 B.xy>0C.x>0,y>0 D.x<0,y<0解析:∵4x2y2=(2xy)2=|2xy|=-2答案:A★2.23-610A.3+2 B.2+3C.1+22 D.1+23解析:原式=23-610-4答案:A3.已知二次函數(shù)y=ax2+bx+0.1的圖象如圖所示,則4(a-A.a+b B.-(a+b)C.a-b D.b-a解析:由圖象知a(-1)2+b×(-1)+0.1<0,∴a<b,∴4(a答案:D4.(a-bA.0 B.2(a-b)C.0或2(a-b) D.a-b答案:C5.已知a∈R,n∈N*,給出下列4個式子:①6(-2)2n;②5a2;③6(-3解析:①中,根指數(shù)為6,是偶數(shù),而被開方數(shù)(-2)2n>0,∴6(-②中,根指數(shù)為5,故5a③中,根指數(shù)為6,是偶數(shù),而被開方數(shù)(-3)2n+1<0,故6(-④中,根指數(shù)為9,故9-a答案:③6.若(42a-1)4+13(a解析:由已知得2a-1≥0,a-3≠0答案:a7.寫出訪下列等式成立的x的取值范圍:(1)31(2)(x-5)(x2解:(1)要使31x-33=1x-(2)(x要使(x-5)2(只需x+5≥0,x-5≤0,★8.已知a<b<0,n>1,且n∈N*,化簡n(解:∵a<b<0,∴a-b<0,a+b<0