由前三項(xiàng)積為3a1*a2*a3=3a1*a1q*a1q^2=3a1^3q^3=3.1后三項(xiàng)積為9a(n-2)a(n-1)an=9a1q^(n-3)a1q^(n-2)a1q^(n-1)=9a1^3q^(3n-6)=9.21式乘2式得a1^3q^3*a1^3q^(3n-6)=27a1^6q^(3n-3)=27[a1^2q^(n-1)]^3=3^3a1^2q^(n-1)=3...某班要從3名男生和3名女生中選出3人分別擔(dān)任數(shù)學(xué)。物理?；瘜W(xué)課代表。要求至少有一名女生。則不同的選法有19若定義在R上的函數(shù)f(x)滿足對(duì)任意x,y屬于R，有f(x+y)=f(x)+f(y)+2.則下列說(shuō)法一定正確的事A.fx是奇函數(shù)B.fx是偶函數(shù)C.fx+2是奇函數(shù)D.fx+2是偶函數(shù)C.fx+2是奇函數(shù)對(duì)任意x1x2∈R有f(x1+x2)=f(x1)+f(x2)+2取x1=x2=0f(0)=2f(0)+2, f(0)=-2取x1=-x2=xf(0)=f(x)+f(-x)+2=-2f(-x)=-4-f(x)設(shè)g(x)=f(x)+2g(-x)=f(-x)+2=-4-f(x)+2=-2-f(x)=-g(x)