廣西專版2024-2025學(xué)年新教材高中數(shù)學(xué)第4章指數(shù)函數(shù)與對數(shù)函數(shù)4.1指數(shù)課后訓(xùn)練新人教A版必修第一冊

4.1指數(shù)課后·訓(xùn)練提升基礎(chǔ)鞏固1.(多選題)下列各式中肯定成立的有()A.(nm)7=n7m17C.4x3+y4=(答案BD2.化簡3-a·A.-a B.--C.-a D.答案A解析明顯a≥0.∴3-a·6a=-a133.當(dāng)a>0時,-ax3=A.xax B.x-C.-x-ax D.-x答案C解析∵-ax3中,∴由a>0得x3≤0,即x≤0,∴-ax3=-4.(3-2x)-34中xA.(-∞,+∞) B.-C.-∞,3答案C解析(3-2x)-34=1(3-5.化簡(a3b12)12÷(a12A.a B.b C.ab D.答案A解析(a3b12)12÷(a12b14)=(a36.若a+b=m13,ab=16m23(m>0),則a3A.0 B.m2 C.-m2 D答案B解析a3+b3=(a+b)(a2-ab+b2)=(a+b)·[(a+b)2-3ab]=m13·(m23-7.化簡(1-2x)2x>1答案2x-1解析∵x>12,∴1-2x<∴(1-2x)2=|1-8.化簡3(-6)3答案-6解析∵3(-6)3=-6,4(5-4)4∴原式=-6+4-5+5-4=-9.若10x=3,10y=4,則102x-y=.

答案9解析∵10x=3,∴102x=9,∴102x-y=1010.化簡下列各式(式中字母均為正數(shù)):(1)b3(2)4x14-3x14y解(1)b3a(2)4x14(-3x14y-13)÷(-6x-11.已知x+y=12,xy=9,且x<y,求:(1)x1(2)x1(3)x-y.解(1)∵(x12+y12)又x12+y12>0,(2)(x12-y12)∵x<y,∴x12-(3)x-y=(x12)2-(y12)2=(x12+y12)(x12-實力提升1.代數(shù)式x-2x恒等于(A.2x3 BC.--2x3 D答案C解析由題知-2x≥0,∴x≤0,∴x-2x=--22.當(dāng)2-x有意義時,化簡x2A.2x-5 B.-2x-1C.-1 D.5-2x答案C解析當(dāng)2-x有意義時,x≤2,則x2-4x+4=(x-2)∴x2-4x+4-x2-6x3.若69a2-6a+1A.(-∞,3) B.-C.13,+答案B解析∵69a2-6a+1=6(1-3a)2=|314.若x<0,則|x|-x2+x2答案1解析∵x<0,∴原式=-x-(-x)+-x-x=-x+x+15.若a>0,且ax=3,ay=5,則a2x+y答案95解析a2x+y2=(ax)2·(ay)12=36.已知3a2+b=1,則9a·答案3解析因為3a2+b=1,所以9a·3b7.已知2a·3b=2c·3d=6,求證:(a-1)(d-1)=(b-1)·(c-1).證明∵2a·3b=6,∴2a-1·3b-1=1.∴(2a-1·3b-1)d-1=1,即2(a-1)(d-1)·3(b-1)(d-1)=1.①又2c·3d=6,∴2c-1·3d-1=1.∴(2c-1·3d-1)b-1=1,即2(c-1)(b-1)·3(d-1)(b-1)=1.②由①②知2(a-1)(d-1)=2(c-1)(b-1),∴(a-1)(d-1)=(b-1)(c-1).8.已知函數(shù)f(x)=x13-x-135(1)求證:f(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞增;(2)分別計算f(4)-5f(2)g(2)和f(9)-5f(3)g(3)的值,由此概括出涉及函數(shù)f(x)和g(x)對全部不等于零的實數(shù)x都成立的一個等式,并加以證明.(1)證明設(shè)x1>x2>0,∵y=x13在R上是增函數(shù),∴又(x1x2)-1∴f(x1)-f(x2)=15(x113-x1-13-x213∴函數(shù)f(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞增.(2)解經(jīng)計算知f(4)-5f(2)g(2)=0,f(9)-5f