21.解:(1)函数f(x)=lm(x+a)+42的定义域为{x1x>-a且x≠-2),且f(x)=x+a(x+2)-(x+a(x+2+4-4a1分讨论:当4-4a≥0时,a≤1,-a≥-1,此时函数f(x)的定义域为(-a,+∞),函数f(x)在区间(-a,+∞)上单调递增;···“···········…………·……······…·················2分当4-4a<0时,a>1.令f(x)>0,解得x>2√a-1或x<-2√a-1(-2√a-1)-(-a)=a-2√a-1=(a-1)-2√a-1+1=(√a-1-1)2.3分1.当a=2时,f(x)=ln(x+2)+4,f(x)=(x+2)x+2函数f(x)的定义域为(-2,+∞).令f(x)=0,得x=2,分析知,函数f(x)在区间(-2,2)上单调递减,在区间(2,+∞)上单调递增…………“·……···········4分.当1 -a>-2,此时函数f(x)的定义域(-a,+∞).令f(x)>0可得-a 2√a-1,所以函数f(x)在区间(-a,-2√a-1)上单调递增,在区间(2√a-1,+∞)上单调递增,在区间(-2√a-1,2√a-1)上单调递减5分ⅲ.当a>2时,-a<-2,此时函数f(x)的定义域为(-a,-2)U(-2,+∞)2√a-1>2,-2√a-1<-2,-a<-2√a-1.分析知,函数f(x)在区间(-a,-2√a-1)上单调递增,在区间(2√a-1,+∞)上单调递增,在区间(-2,2√a-1)上单调递减,在区间(-2√a-1,-2)上单调递减6分(2)由(1)求解知,当1 8-2n2,所以0<1 <分,得0<2-a<,即 8-22,则实数a的取值范围为(2,2)………………………12分

书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!

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