Question

求此数学题详细解题过程。。。。。。。。。。。。。

课本最后答案是这个，求过程，不理解怎么来的这个解

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Answer 1

(ab^-1c^3)的根号÷(a^2b^-2c^2)的3次方根

a,b,c 都不等于0.
[ab^(-1)c^3] > 0.

原式 = [ab^(-1)c^3]^(1/2)/[a^2b^(-2)c^2]^(1/3)

= [ab^(-1)c^3]^(1/2)c^(4/3)/{[a^2b^(-2)c^2]^(1/3)*c^(4/3)}

= [ab^(-1)c^3]^(1/2)c^(4/3)/[a^2b^(-2)c^2*c^4]^(1/3)

= [ab^(-1)c^3]^(1/2)c^(4/3)/[a^2b^(-2)c^6]^(1/3)

= [ab^(-1)c^3]^(1/2)c^(4/3)/[ab^(-1)c^3]^(2/3)

= [ab^(-1)c^3]^[1/2 - 2/3]c^(4/3)

= [ab^(-1)c^3]^[-1/6]c^(4/3)

Answer 2

原式=a^(1/2)b^(-1/2)c^(3/2)÷[a^(2/3)b^(-2/3)c^(2/3)]
=a^(1/2-2/3)b^(-1/2+2/3)c^(3/2-2/3)
=a^(-1/6)b^(1/6)c^(5/6)

Answer 3

按你的写法这样做比较合适

你说的课本答案和我的一样，你把a当作a的六次方再开6次方

Answer 4

这种题还是要有用笔算