Question

2乘以根号3乘以三次根号6乘以六次根号12除以（三次根号4乘以三次根号23）

快快！
我打错了最后是三次根号27

Answer 1

2乘以根号3乘以三次根号6乘以六次根号12除以（三次根号4乘以三次根号3）??
=2*3^(1/2)*6^(1/3)*12^(1/6)/[4^(1/3)*3^(1/3)]
=2*3^(1/2)*(2*3)^(1/3)*(2*2*3)^(1/6)/[(2*2)^(1/3)*3^(1/3)]
=2*3^(1/2)*2^(1/3)*3^(1/3)*2^(1/6)*2^(1/6)*3^(1/6)/[2^(2/3)*3^(1/3)]
=2^(1+1/3+1/6+1/6-2/3)*3^(1/2+1/3+1/6-1/3)
=2*3^(2/3)
=2乘以三次根号(9)

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我打错了最后是三次根号27

追答

2乘以根号3乘以三次根号6乘以六次根号12除以（三次根号4乘以三次根号27）
=2*3^(1/2)*6^(1/3)*12^(1/6)/[4^(1/3)*3]
=2*3^(1/2)*(2*3)^(1/3)*(2*2*3)^(1/6)/[(2*2)^(1/3)*3]
=2*3^(1/2)*2^(1/3)*3^(1/3)*2^(1/6)*2^(1/6)*3^(1/6)/[2^(2/3)*3]
=2^(1+1/3+1/6+1/6-2/3)*3^(1/2+1/3+1/6-1)
=2*3^(0)
=2*1
=2

Answer 2

最后是24吧

原式=2*3^(1/2)*6^(1/3)*12^(1/6)÷(4^(1/3)*24^(1/3)
=2*3^(1/2)*2^(1/3)*3^(1/3)*2^(1/3)*3^(1/6)÷(2^(2/3)*2*3^(1/3))
=2^(1+1/3+1/3-2/3-1)*3^(1/2+1/3+1/6-1/3)
=2^0*3^(2/3)
=3次根号9

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我打错了最后是三次根号27

Answer 3

原式=2根号3*6^(1/3)*12^(1/6)/[4^1/3*23^1/3]
=2*3^1/2*(2*3)^1/3*(2^2*3)^1/6/[(2^2)^1/3*23^1/3]
=2^(1+1/3+1/3)*3^(1/2+1/3+1/6)/[2^(2/3)*23^1/3]
=2^(5/3-2/3)*3/23^1/3
=2*3/23^1/3
=6/(三次根号23)

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我打错了最后是三次根号27

追答

原式=2根号3*6^(1/3)*12^(1/6)/[4^1/3*27^1/3]
=2*3^1/2*(2*3)^1/3*(2^2*3)^1/6/[(2^2)^1/3*(3^3)^1/3]
=2^(1+1/3+1/3)*3^(1/2+1/3+1/6)/[2^(2/3)*3]
=2^(5/3-2/3)*3/3
=2*3/3
=2

Answer 4

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我打错了最后是三次根号27

追答

23算了很久都算不出来

其实刚才把第一张图片发给你的时候就已经把思路发给你了

求采纳