利用幂的运算性质计算

√18÷√3
=√3*√6÷√3
=√6

5次根号4乘根号8除以10次根号16
=4^1/5*8^1/2÷16^1/10
=(2^2)^1/5*(2^3)^1/2÷(2^4)^1/10
=2^2/5*2^3/2÷2^2/5
=2^(2/5+3/2-2/5)
=2^(3/2)
=2√2

3次根号3除以3次根号-9分之1乘3次根号3的5次方
=3^1/3÷(-1/9)^1/3*(3^5)^1/3
=3^1/3÷[-(1/3)^2]^1/3*(3^5)^1/3
=3^1/3÷[-(1/3)^2/3]*3^5/3
=3^1/3÷[-3^(-2/3)*3^5/3
=-3^1/3÷3^(-2/3)*3^5/3
=-3^[1/3-(-2/3)+5/3]
=-3^[1/3+2/3+5/3]
=-3^(8/3)

4次根号3乘根号12
=3^1/4*12^1/2
=3^1/4*(3*2^2)^1/2
=3^1/4*3^1/2*2
=2*3^(1/4+1/2)
=2*3^(3/4)

3次根号3乘3次根号18
=3^1/3*18^1/3
=(3*18)^1/3
=(2*3^3)^1/3
=2^1/3*3^(3*1/3)
=3*2^(1/3)

2的3分之4次方除以3的3分之2次方乘6的3分之2次方（结果表示为含幂的形式）
=2^(4/3)÷3^(2/3)*6^(2/3)
=2^(4/3)÷3^(2/3)*(2*3)^(2/3)
=2^(4/3)÷3^(2/3)*2^(2/3)*3^(2/3)
=2^(4/3)*2^(2/3)*3^(2/3)÷3^(2/3)
=2^(4/3+2/3)*3^(2/3-2/3)
=2^2*3^0
=2^2
=4

6的4分之1次方乘12的4分之3次方（结果表示为含幂的形式）
=6^(1/4)*12^(3/4)
=6^(1/4)*(2*6)^(3/4)
=6^(1/4)*2^(3/4)*6^(3/4)
=6^(1/4)*6^(3/4)*2^(3/4)
=6^(1/4+3/4)*2^(3/4)
=6*2^(3/4)

（2分之1）的2分之1次方乘（2分之3）的-2分之1次方乘（4分之3）的2分之1次方
=(1/2)^1/2*(3/2)^(-1/2)*(3/4)^(1/2)
=(1/2)^1/2*(2/3)^(1/2)*(3/4)^(1/2)
=(1/2*2/3*3/4)^(1/2)
=(1/4)^(1/2)
=[(1/2)^2]^(1/2)
=(1/2)^(2*1/2)
=1/2

√18/√3=√(18/3)=√6
5√4*√8/10√16=2＾(2/5)*2＾(3/2)*2＾(4/10)=2＾(2/5+3/2+4/10)=2＾(2+3/10)=4*2＾(3/10)
3√3/1/[3√(-9)]*3√(3＾5)=3＾(1/3)*[-3＾(3/2)]*3＾(5/3)=-3＾(1/3+3/2+5/3)=-3＾(3+1/2)=-27*√3
4√3*√12=3＾(1/4)*2*√3=2*3＾(3/4)
3√3*3√18=3√(3*18)=3√54=3*2＾(1/3)
2＾(4/3)/4＾(2/3)*6＾(2/3)=2＾(4/3+4/3+2/3)*3＾(2/3)=8*2＾(1/3)*3＾(2/3)
6^(1/4)*12^(3/4)=2^(1/4+6/4)*3^(1/4+3/4)=2*2^(3/4)*3=6*2^(3/4)
(1/2)^(1/2)*(2/3)^(-1/2)*(3/4)^(1/2)=2^(-1/2)*2^(-1/2)*3^(1/2)*3^(1/2)*2=2*2^(-1)*3^(1)=3

根号6
2倍根号2
-3倍根号3
2乘以3的3/4次幂
3乘以3次根号2
4倍3的4/3次幂
3倍2的7/4次幂
1/2

先把化简得到原式,然后根据分数指数的意义得到原式,再利用积的乘方得到原式,然后再写成根式的形式.
解:原式.
本题考查了分数指数幂:(与都为正整数).