Loop decimal variable fraction solution 0 3544 method?

The solution of cyclic decimal fractions requires the knowledge of the summation formula of the proportional sequence and the limit.

To put it simply, if the cyclic decimal is regarded as the summation, where it is a constant, and secondly, if it is regarded as the first term of the proportional sequence a1, and the common ratio q=, then the infinite cyclic decimal can be regarded as the sum of the proportional series, which is calculated by the formula as.

sn=a1×(1-q

n)/(1-q)

In the above equation, when n tends to infinity, it tends to 0, and sn=then.

The answer is 319 900, or 315 out of 900

Let x=, the loop section is 4, and there is only 1 digit.

10x=,10x-x=,9x=,x=,so the cyclic decimal becomes a fraction,.

x=10x=

10x-x=9x=

x= similar, multiply by 100 and subtract.

The original number contains cavity 10 + 1 22 = 19 55

This problem is used: sum of infinite sequences: s=a1 (1-q) (the formula is to sum the cyclic body).

where a1 is the first of the cycle, i.e.

where q is the common ratio: i.e., the ratio of the second cycle to the first Sennai cycle:

Original Numbers This question uses a cavity: sum of infinite sequences: s=a1 (1-q) (the formula for talking about spring shirts is to sum the cyclic body).

where a1 is the first of the cycle, i.e.

where q is the common ratio: i.e., the ratio of the second cycle to the first cycle:

Summary. The cyclic decimal fraction is 4 and 1 3... The cyclic decimal fraction of 1 is 1 to 3 cyclic decimal fraction.

The cyclic decimal fraction is 4 and 1 3... The cyclic decimal fraction is 1 3 and the cyclic decimal is x. 10x-x=39 can be obtained, and the equation can be solved to get x=4 and 1 3

Pure cyclic decimal fractions, with the denominator written as 9 and the cyclic number 3 written as a continuum. Re-reduction can also be turned into fractions.

Teach you a mantra to take out the integer part first, leaving only the decimal part.

Then, the denominator is: a few cycles and a few nines, and a few non-cycles and a few 0 numerators are: the fractional part of all the base silver (the cycle is counted once) minus the part that is not cyclical.

For example, a cyclic knot (2) is a 9, and two non-cyclic (43) is two 0s, and the denominator is 900

The whole part 432 minus the non-circular part 43 equals 389, so the result is 389 900, and if there is an integer part, it is enough to add the integer part

Remember...

, 53 loop (the loop above 53 after the decimal point cannot be typed) set, then 100x=, so there is.

100x-x=

99x=53

So x is equal to 53/99, ie.

The end of the cyclic decimal is 3 times 3, so multiply the cyclic by 3, equal, approximately equal, equal to 5 8, and then multiply 5 by 3 equals 15, so the result is 15 8

Putting the part of the loop (assuming there is an n-bit loop) as the numerator, and taking n 9s as the denominator, the resulting fraction is the simplest fraction form of the decimal number.

For example: =1 9

And so on ......

It depends on whether the cycle section is 25 or 5:

If the cycle section is 25, then 25 out of 25

If the cycle section is 5, then it is 23 out of 23

It is a pure cyclic decimal, and when the pure cyclic decimal is converted into a fraction, there are several cyclic node denominators with several 9s, and the numerator is a cyclic node, if the denominator and the numerator can be reduced to a recurrent simple fraction.

It is a two-digit cyclic node, the denominator is 99, the numerator is 25, and the fraction is 25/99 is a mixed cyclic decimal, and the process of fraction is as follows

1/1 + 5/90 = 23/90