Explore Fraction Concepts with Examples and Questions

Is there a formula to approximate $\mathrm{\pi <!--\; \pi \; -->}$in the form of ${\displaystyle \frac{p}{q}}$?

Adding fractions with variables and using common denominator? Rather basic math, but here I am..
So I have these fractions I need to merge together and shorten:
$\frac{1}{2a+8}+\frac{4}{a^2-<!--\; -\; -->16}+\frac{4}{a-<!--\; -\; -->4}$
I understand that I need to make the denominators the same so that I can merge these fractions together, but I have no idea how.
I assume I need to find the common denominator, which is a friend told me was $2a^2-<!--\; -\; -->64$. I don't know what to do from here, please help?
How do I use the common denominator to make these denominators the same?
EDIT: Okay so I failed miserably at this MathJax thing. What I am trying to say is:
$\frac{1}{2a+8}+\frac{4}{a^2-<!--\; -\; -->16}+\frac{4}{a-<!--\; -\; -->4}$
A quick lesson in writing fractions on this forum would also be immensely appreciated.

If $\frac{a}{b}=\frac{x}{y}$, is $\frac{x-<!--\; -\; -->a}{y-<!--\; -\; -->b}=\frac{x}{y}$?

When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$?
The solution to the quadratics is given by
$r=-<!--\; -\; -->{\displaystyle \frac{b}{2a}}\pm <!--\; \pm \; -->\sqrt{{\displaystyle \frac{b^2-<!--\; -\; -->4ac}{4a^2}}}$, which is shortened to $r=-<!--\; -\; -->{\displaystyle \frac{b}{2a}}\pm <!--\; \pm \; -->{\displaystyle \frac{\sqrt{b^2-<!--\; -\; -->4ac}}{2a}}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a\in <!--\; \in \; -->\mathbb{C}$, and $\sqrt{{\displaystyle \frac{x}{y}}}\ne <!--\; \ne \; -->{\displaystyle \frac{\sqrt{x}}{\sqrt{y}}}$ if x and y are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2}=\sqrt{x}/y$? Is $r=-<!--\; -\; -->{\displaystyle \frac{b}{2a}}\pm <!--\; \pm \; -->{\displaystyle \frac{\sqrt{b^2-<!--\; -\; -->4ac}}{2a}}$ always true?

fraction simplication, resolution with one variable
I'm having some problems to understand a simplification and I know the correct answer but I'd like to understand why my method is wrong or what rule I'm breaking.
I have an equation like the following:
$Fr\cdot <!--\; \cdot \; -->(\frac{1}{Sr}+\frac{1}{Sp})=\frac{Ft}{Sp}$
Where the variable I'm looking for is $Fr$. My tendency was to do:
$Fr=\frac{Ft}{Sp}\cdot <!--\; \cdot \; -->(Sr+Sp)$
But it seems to be wrong, I should have do:
$Ft\cdot <!--\; \cdot \; -->\left(\frac{Sp+Sr}{Sr\cdot <!--\; \cdot \; -->Sp}\right)=\frac{Ft}{Sp}$
and then:
$Ft=\frac{Sr\cdot <!--\; \cdot \; -->Sp\cdot <!--\; \cdot \; -->Ft}{Sp+Sr}$
Why I can not simply pass & inverse the fraction to the other side? Thanks so much!!

Ratio of sums vs sum of ratio
Is anyone aware of any general (or perhaps not so general) relationship (inequality for instance) relating
$A(x,y)=\frac{\underset{z}{\sum <!--\; \sum \; -->}f(x,y,z)}{\underset{z}{\sum <!--\; \sum \; -->}g(y,z)}$
and
$B(x,y)=\underset{z}{\sum <!--\; \sum \; -->}\left(\frac{f(x,y,z)}{g(y,z)}\right)$
?
Specific context (for what I'm dealing with, but not necessarily the question) is that $\underset{x,y,x}{\sum <!--\; \sum \; -->}f(x,y,z)=1$ and $\underset{y,x}{\sum <!--\; \sum \; -->}g(y,z)=1$ and $f(x,y,z)\ge <!--\; \ge \; -->0$ and $g(y,z)\ge <!--\; \ge \; -->0\mathrm{\forall <!--\; \forall \; -->}x,y,z$. I.e. probabilities (or more generally, I guess, measures).
It seems like it could 'vaguely' be related to log sum inequalities (when transformed) or Jensen's inequality perhaps?

Working out cost based on time spent - simple math
I did a task, and my hourly rate is $\pounds 25$ , I spent a total of $36$ minutes on it, how can I work out the total amount of time spent on the task?
My attempt:
I can do this for simple sums such as $30$ mins $=25/2=\pounds 12.5⟹<!--\; ⟹\; -->25$ mins $=25/4=\pounds 6.25$ but finding it hard for numbers such as $36$ mins.
Thank you

Sign of fractional exponent
What is the sign of $-<!--\; -\; -->1^{\frac{2}{3}}$? I thought it was positive 1 because it involves squaring, but that doesn't seem to be the case. Why?

Integrate algebraic fraction with constant on top?
I understand that if you have $\int <!--\; \int \; -->\frac{1}{x+1}dx$ you simply do $\ln <!--\; \; -->(x+1)+C$. Now I'm slight confused because in my text book, $\int <!--\; \int \; -->\frac{31}{x-<!--\; -\; -->4}dx$ evaluates to $31\ln <!--\; \; -->(x-<!--\; -\; -->4)$ but $\int <!--\; \int \; -->\frac{-<!--\; -\; -->2}{4x-<!--\; -\; -->1}dx$ becomes $-<!--\; -\; -->\frac{1}{2}\ln <!--\; \; -->(4x-<!--\; -\; -->1)$. Why is this?

Simplify $\frac{5x}{x^2-<!--\; -\; -->x-<!--\; -\; -->6}+\frac{4}{x^2+4x+4}$
How come the answer is left as $\frac{5x}{(x+2)(x-<!--\; -\; -->3)}+\frac{4}{(x+2{)}^2}$. Why don't we go any further?

How does $-<!--\; -\; -->\sqrt{\frac{2-<!--\; -\; -->\sqrt{2}}{2+\sqrt{2}}}$ simplify to $1-<!--\; -\; -->\sqrt{2}$
I've the answer for a question in my textbook to be: $-<!--\; -\; -->\sqrt{\frac{2-<!--\; -\; -->\sqrt{2}}{2+\sqrt{2}}}$ which i've then simplifed to: $-<!--\; -\; -->\sqrt{3-<!--\; -\; -->2\sqrt{2}}$
However my textbook states $-<!--\; -\; -->\sqrt{\frac{2-<!--\; -\; -->\sqrt{2}}{2+\sqrt{2}}}=1-<!--\; -\; -->\sqrt{2}$
How is this?

Is this a mistake on my part or theirs?
I'm not sure if I'm the one making the mistake, or my math book. It looks like the negative sign completely disappeared.
$\frac{3x^2}{-<!--\; -\; -->\sqrt{18}}\cdot <!--\; \cdot \; -->\frac{\sqrt{2}}{\sqrt{2}}=\frac{3x^2\sqrt{2}}{-<!--\; -\; -->\sqrt{36}}=\frac{3x^2\sqrt{2}}{6}=\frac{x^2\sqrt{2}}{2}$
Also, I am new to Stackexchange, so tell me if I am doing something wrong.

How is $\frac{({10}^4{)}^6-<!--\; -\; -->1}{{10}^4-<!--\; -\; -->1}=1+{10}^4+{10}^8+{10}^{12}+{10}^{16}+{10}^{20}$?

Algebraic fractions: addition
I have very elementary question about adding algebraic fractions. Now, I know the following:
$\frac{a}{c}+\frac{b}{d}=\frac{da+cb}{dc}$
My question is however, how given this expression:
$\frac{(2n+1)n(n+1)}{2}+\frac{(2n+1)n(n+1)}{6}$
One arrives at:
$\frac{(2n+1)n(n+1)}{3}$
Trying to work out the solution, I arrive at a very large equation divided by 12...

Estimating the natural logarithm from both sides:
$1/(a+1)<\ln <!--\; \; -->((1+a)/a)<1/a$
I must prove that for all $a>0$
$\frac{1}{a+1}<\ln <!--\; \; -->\frac{1+a}{a}<\frac{1}{a}$
Can anyone help me?

Repeating decimal notation of 1/53 on WolframAlpha vs notation on Wikipedia
WolframAlpha shows for 1/53
$0.0\stackrel{¯<!--\; ¯\; -->}{1886792452830}$
as the repeating decimal. Why is it not
$0.\stackrel{¯<!--\; ¯\; -->}{0188679245283}$ instead?
For example, Wikipedia shows for 1/81
$0.\stackrel{¯<!--\; ¯\; -->}{012345679}$
as the repeating decimal, not
$0.0\stackrel{¯<!--\; ¯\; -->}{123456790}$

If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2{c}^2+a^3{c}^2}{b^{4}c+d^4+b^{2}c{d}^2}}$
What I've done so far;
$$\begin{gathered} \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k \\ a=bk,b=ck,c=dk \\ a=c{k}^2,b=d{k}^2 \\ a=d{k}^3 \end{gathered}$$
I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.

Distribution of the sum reciprocal of primes $\le <!--\; \le \; -->1$
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\cdots <!--\; \cdots \; -->\le <!--\; \le \; -->1$
This is an interesting infinite summation. This is very closely resembling my other problem with has to do with the distribution of composites. That other problem is here. The same rules of that other summation apply, but it's good to repeat them :).
Rules of this summation:
1) All terms of the summation must be reciprocal of some prime. For example, $\frac{1}{6}$ could never be a term because 6 is not a prime number.
2) The next term must be the largest reciprocal term that does not allow the entire sum to be $>1$. The next term also must be < the term before it.
3) The sum must also be permanently less than 1 if you were to stop at a number $<<!--\; <\; -->\mathrm{\infty <!--\; \infty \; -->}$. If you were to stop on term $\mathrm{\infty <!--\; \infty \; -->}$, then the sum would amount to exactly 1 by definition.
Now that you know the rules, we must rewrite the summation in a different way in order to propose my question in a reasonable manner.
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\cdots <!--\; \cdots \; -->\le <!--\; \le \; -->1$
As a mote of clarification, $a,b,c,d$ etc. represent $2,3,7,43$ etc. respectively.
The question:
-Envision $f(1)=a,f(2)=b,f(3)=c$, and so on. Keep in mind there are an infinite number of terms, thus an infinite number of variables to represent primes. My question is what function, $f(n)$, could hold this property and maintain true no matter what term you are at.
-One who is loving this must remember that $a,b,c,d$, etc. are integers rather than fractions in this case and therefor $f(1)\ne 1/a,f(1)\ne 1/b$, etc. The reciprocal however is the form in which the numbers have that interesting property.
-What is $f(n)$? (The primary question)
-When answering this question, do the following, please provide some form of evidence or explain your case. If I missed something obvious, please let me know what that is with a website in the comments.
Have a great day! Thanks!

Equation simplification, can't get it right
$\frac{1}{\frac{x-<!--\; -\; -->1}{x+2}}-<!--\; -\; -->\frac{2}{x^2-<!--\; -\; -->1}$
should be simplified into
$\frac{x^2+3x}{x^2-<!--\; -\; -->1}.$
However, when I try to do it (tried several times), I fail to get it done right:
$\frac{1}{\frac{x-<!--\; -\; -->1}{x+2}}-<!--\; -\; -->\frac{2}{x^2-<!--\; -\; -->1}=1\ast <!--\; \ast \; -->\frac{x+2}{x-<!--\; -\; -->1}-<!--\; -\; -->\frac{2}{x^2-<!--\; -\; -->1}$
$=\frac{x(x+2)-<!--\; -\; -->2}{x^2-<!--\; -\; -->1}=\frac{x^2+2x-<!--\; -\; -->2}{x^2-<!--\; -\; -->1}$
What am I doing wrong?
Thanks a lot for the help

Find the water pumped to flowers using fractions.
A certain tank has water filled to $\frac{1}{5}$ of it's capacity. A persons opens a tap and allows the tank to fill upto to $\frac{3}{4}$ of it's capacity. Then he uses the water in the tank and finally $\frac{1}{3}$ water remains in the tank.
What I have found :
Water that was added to the tank after opening the tap : $\frac{11}{20}$
Fraction of water he used was : $\frac{5}{12}$
Total capacity of water in the tank : 1800 litres
Problem:
Find the amount water he used up in litres
How many more litres remained in the tank in the end when compared to the beginning