开平方的简化

正确选项是？

import numpy as np
from latexifier import latexify
from IPython.display import Latex
import fractions


s = '√3√3√3√3'
# Numerator, denominator 分子，分母

def recur(s):
    d = int(s[-1])
    nd = 0
    for c in s[::-1]:
        print(nd,c,s[-1],c==s[-1])
                
        if c == s[-1]:
            nd += 1
        
        elif c == '√':
            #n = n**0.5
            nd = 0.5*nd

    return fractions.Fraction(nd)
    
print(recur(s))
15/16

n,d = str(recur(s)).split("/")
print(n,d)
15 16

display(Latex('$$ (√3√3√3√3) = 8 ^ {\dfrac{15}{16}} $$'))

输出格式

正确选项是 C

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