使用the decimal module：

import decimal
D = decimal.Decimal
n = D(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)

with decimal.localcontext() as ctx:
    ctx.prec = 300
    x = n.sqrt()
    print(x)
    print(x*x)
    print(n-x*x)

收益率

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999983754999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998612677

99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

0E-100

math.sqrt返回IEEE-754 64位结果，大约为17位，还有其他库可以处理高精度值，除了上面提到的decimal和mpmath库之外，我还维护gmpy2库（https://code.google.com/p/gmpy/）。

>>> import gmpy2
>>> n=gmpy2.mpz(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
>>> gmpy2.get_context().precision=2048
>>> x=gmpy2.sqrt(n)
>>> x*x
mpfr('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0',2048)
>>>

gmpy2库还可以返回整数平方根（isqrt）或快速检查整数是否为精确平方（is_square）。

这是一个整数平方根程序，使用了我不久前写的Hero方法，初始近似值是输入值位长的一半，所以收敛速度很快，但是，我还没有计时，看看在Python中它是否比使用更简单的初始近似值更快。

#! /usr/bin/env python

''' Long integer square roots. Newton's method.

    Written by PM 2Ring. Adapted from C to Python 2008.10.19
'''

import sys

def root(m):
    # Get initial approximation
    n, a, k = m, 1, 0
    while n > a:
        n >>= 1
        a <<= 1
        k += 1
        #print k, ':', n, a

    # Go back one step & average
    a = n + (a>>2)
    #print a

    # Apply Newton's method
    while k:
        a = (a + m // a) >> 1
        k >>= 1
        #print k, ':', a
    return a

def main():
    m = len(sys.argv) > 1 and int(sys.argv[1]) or 2*10L**100
    print "The Square Root of", m
    print root(m)

if __name__ == '__main__':
    main()

你可以用python3 Num类（包num7）找到非常大的数的根，如下所示：

#square root of very large numbers
from num7 import Num

n = Num('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0')
r1 = n.sqrt()
print('n=>', n)
print('DIGITS', n.len()) #DIGITS (200, 0)
print('r1=>', r1)#9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999
print('DIGITS', r1.len())#DIGITS (100, 80)
print('--'*20) 

x2 = r1*r1     
print('x2=>', x2)#99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982919999999999999999999999999999999999999999999999999999999999999999999999999999999800000000000000729316.0000000000000000000000000000000000000000000000000000000000000000000000000000170800000000000000000000000000000000000000000000000000000000000000000000000000000001
print('DIGITS', x2.len())#DIGITS (200, 160)
print('--'*20)
print('difference n-x2=>', n-x2) #199999999999999996750.9999999999999999999999999999999999999999999999999999999999999999999999999999829199999999999999999999999999999999999999999999999999999999999999999999999999999999