5.∛(a^(9/2) √(a^("-" 3) )) ÷√(∛(a^("-" 7) ) "·" ∛(a^13 )) =　　　　　(其中a>0).

解析:原式=[a^(1/3×9/2)·a^(1/3×("-" 3/2) )]÷[a^(1/2×("-" 7/3) )·a^(1/2×13/3)]=a^(9/6 "-" 3/6+7/6 "-" 13/6)=a0=1.

答案:1

6.已知a=-27/125,b=(2" " 017)/(2" " 018),试求(a^(2/3)+3∛ab+9b^(2/3))/(a^(4/3) "-" 27a^(1/3) b)÷a^(1/3)/(∛a "-" 3∛b)的值.

解:显然a≠0,所以有

　　原式=(a^(2/3)+3a^(1/3) b^(1/3)+"(" 3b^(1/3) ")" ^2)/(a^(1/3) "(" a"-" 27b")" )×(a^(1/3) "-" 3b^(1/3))/a^(1/3)

　　=("(" a^(1/3) ")" ^3 "-(" 3b^(1/3) ")" ^3)/(a^(2/3) "(" a"-" 27b")" )=1/a^(2/3) =a^("-" 2/3)=("-" 27/125)^("-" 2/3)=25/9.

7.导学号85104060(拓展探究)已知a>0,若对于a≤r≤8,r∈N+,式子(√a)8-r·(1/∜a)^r能化为关于a的整数指数幂的可能情形有几种?

解:(√a)8-r·(1/∜a)^r=a^((8"-" r)/2)·a^("-" r/4)=a^((16"-" 3r)/4).

　　∵a>0,a≤r≤8,r∈N+,

　　∴r=4,8时,上式能化为关于a的整数指数幂,故符合要求的情形有两种.