Explore Fraction Concepts with Examples and Questions

Finding average of denominator knowing average of numerator and average of fraction
Good day to you all. I have a little problem I have been banging my head on for a while.
I have come to think that it is impossible, but I hope you can save me.
I have a fraction, $\frac{nu{m}_i}{de{n}_i}$., which takes different values over time.
The objective is to calculate $\frac{average(nu{m}_i)}{average(de{n}_i)}$
I have at my disposal $average(\frac{nu{m}_i}{de{n}_i})$ and $average(nu{m}_i)$
Is there any way to do this, or do I need to get $average(de{n}_i)$ also?
Thanks a lot for your help.
Edit with an example
Let's take $\frac{1}{2},\frac{1}{3},\frac{1}{4}$
The information at my disposal is:
The average of numerators is 1.
The average of fractions is 0.36
Is it possible with the information I have to retrieve the average of denominators?

A fraction problem
$$\begin{gathered} a=x+\frac{1}{x} \\ b=y+\frac{1}{y} \\ c=xy+\frac{1}{xy} \end{gathered}$$
Express $c$ in terms of $a$ and $b$

Algebraic Problem Fractions
Given the following fraction,
$\frac{n}{2^{n-<!--\; -\; -->1}}(-<!--\; -\; -->2{)}^{n-<!--\; -\; -->1}$
Is it correct that it simplifies to the following fraction,
$(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->1}n$
I am in doubt because I have checked on three different places and they all gave different answer.

Prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$
I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a,b,c,d$ be positive real numbers, prove that if $\frac{a}{b}<\frac{c}{d}$, then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$

How do I simplify $\frac{\sqrt{21}-<!--\; -\; -->5}{2}+\frac{2}{\sqrt{21}-<!--\; -\; -->5}$

Defining priority of operations in limits with stacking fractions
I need to evaluate the following limit :
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(x^2+1)}{x^2}$
Using L'Hospital rule, I get this result (which I'm pretty sure is good)
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\frac{2x}{x^2+1}}{2x}$
Now, I'm not sure how I must evaluate this. Either as :
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2x}{x^2+1}\ast <!--\; \ast \; -->\frac{1}{2x}=\frac{1}{x^2+1}$
or
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}2x\ast <!--\; \ast \; -->(\frac{x^2+1}{2x}{)}^{-<!--\; -\; -->1}=\frac{4x^2}{x^2+1}$
According to most of the tools I use to validate my maths, the 1st result is the good one, but I can't figure out why, am I simply missing a set of parenthesis in my equation?

At what $a$ is $\underset{x\to <!--\; \to \; -->0}{lim}\frac{e^{ax}-<!--\; -\; -->e^x-<!--\; -\; -->x}{x^2}$ finite? What is that limit?
Here is a function:
$\frac{e^{ax}-<!--\; -\; -->e^x-<!--\; -\; -->x}{x^2}$
In which $a$ is a coefficient. The problem whats us the value for $a$ which gives a finite value for the limit,
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{e^{ax}-<!--\; -\; -->e^x-<!--\; -\; -->x}{x^2}$
How to find that value? I first though that the degree of the numerator must be equal to that of the denominator. but I don't know what the definition of degree is for $e^x$
Even if I knew, I couldn't find the limit.

Write as a single fraction in its simplest form
So it seems as though am having quite a bit of trouble with this question of mine, i've spent some time trying to find the correct way to do it but am unsure about how to do it since i always seem to end up in a slight muddle i guess.
The question:
Write as a single fraction in its simplest form:
$\frac{4x-<!--\; -\; -->1}{2(x-<!--\; -\; -->1)}$ - $\frac{3}{2(x-<!--\; -\; -->1)(2x-<!--\; -\; -->1)}$
I've tried expanding the brackets first to end up with:
$\frac{4x-<!--\; -\; -->1}{2x-<!--\; -\; -->2}$- $\frac{3}{2x-<!--\; -\; -->2(2x-<!--\; -\; -->1)}$
(All of the signs are negative if you couldn't tell)
And i was then going to combine each of the denominators with their respective numerators (opposite method) and then simplify and go from there...
I've tried different methods too but i dohn't think any of the ones i have tried are accurate and am not entirely sure if this one is either :/
A detailed method would be greatly appreciated but ill be happy if any advice is provided.
Thank a bunch :)

How to solve for $x$ in $\sqrt{9+2x}-<!--\; -\; -->\sqrt{2x}=\frac{5}{\sqrt{9+2x}}$?

$\left[\frac{n}{1}\right]+\left[\frac{n}{2}\right]+\cdots <!--\; \cdots \; -->+\left[\frac{n}{n}\right]+\left[\sqrt{n}\right]$ is even
I tried this by taking two different cases, when, $n$ is even and when $n$ is odd. But I could not establish any relation between the two. Please help to solve this problem and thanks a lot in advance.
P.S. Although nothing is mentioned in the question I think $[]$ stands for greatest integer function as otherwise the sum would be in a fraction which is not even as only natural numbers are regarded as even.

Simplifying radical expression $\frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3q^2+4pq+p^2}}\cdot <!--\; \cdot \; -->\frac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}÷<!--\; ÷\; -->\frac{1}{\sqrt{p^2+3pq+2q^2}}$
Problem:
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3q^2+4pq+p^2}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}}÷<!--\; ÷\; -->{\displaystyle \frac{1}{\sqrt{p^2+3pq+2q^2}}}$
My try:
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}}{\sqrt{3p+q}\sqrt{p+q}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+3q}}{\sqrt{p+4q}\sqrt{p+2q}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{p+2q}\sqrt{p+q}}{1}}=$
${\displaystyle \frac{\sqrt{p-<!--\; -\; -->4q}\sqrt{p+3q}}{\sqrt{3p+q}\sqrt{p+4q}}}=$
${\displaystyle \frac{\sqrt{p^2-<!--\; -\; -->pq-<!--\; -\; -->12q}}{\sqrt{3p^2+13pq+4q^2}}}$
How do I continue?

Question about an inequality which seems right but not easy to prove
The origin problem is as follows:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ prove:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->4+\frac{1}{4}[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2]$
The solution is easy enough, which is to plug $4=a+b+c+d$ into to $RHS$ and then move it to $LHS$, notice:
$[\frac{(a-<!--\; -\; -->b{)}^2}{b}+\frac{(b-<!--\; -\; -->c{)}^2}{c}+\frac{(c-<!--\; -\; -->d{)}^2}{d}+\frac{(d-<!--\; -\; -->a{)}^2}{a}](a+b+c+d)\ge <!--\; \ge \; -->[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2]$
which would finally lead to the proof of the problem.
However, when I tried to solve the problem, I applied the Lagrange identical equation to the $RHS$ of the inequality, then I left out some quadratic term which finally leads to the inequality:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->a^2+b^2+c^2+d^2$
Which seems quite right, but I'm not really sure if it is. However, numerical tests imply that it is right. But I'm still not sure.
That stuff is kind of like Chebyshev inequality but it can't be directly used here.
From another point of view, my question is whether the following strengthening　of the origin inequality is right or not:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ then:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge <!--\; \ge \; -->4+\frac{1}{4}[(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2+(a-<!--\; -\; -->c{)}^2+(b-<!--\; -\; -->d{)}^2]$

simplify fractions with exponent
The given fraction is: $(\frac{a}{b}{)}^n\cdot <!--\; \cdot \; -->(\frac{b}{c}{)}^n\cdot <!--\; \cdot \; -->(\frac{c}{a}{)}^{n+1}$
The given solution is: $\frac{c}{a}$
What I have done so far:
$(\frac{a}{b}{)}^n\cdot <!--\; \cdot \; -->(\frac{b}{c}{)}^n\cdot <!--\; \cdot \; -->(\frac{c}{a}{)}^{n+1}$ | multiply $\frac{a}{b}$ and $\frac{b}{c}$ because of same exponent
$(\frac{ab}{bc}{)}^n\ast <!--\; \ast \; -->(\frac{c}{a}{)}^{n+1}$ | get rid of $b$
$(\frac{a}{c}{)}^n\ast <!--\; \ast \; -->(\frac{c}{a}{)}^{n+1}$
Can you please explain how I continue simplifying or what I did wrong? Thanks!

How do I solve this, first I have to factor $\frac{2x}{x-<!--\; -\; -->1}+\frac{3x+1}{x-<!--\; -\; -->1}-<!--\; -\; -->\frac{1+9x+2x^2}{x^2-<!--\; -\; -->1}$
the solution is
$\frac{3x}{x+1}$
The only advance that I have done is factor $x^2-<!--\; -\; -->1$=$(x-<!--\; -\; -->1)$$(x+1)$
do not know how can I factor $1+9x+2x^2$ can someone please guide me in how to solve this exercise.

when the sum of some fractions be $1$
prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.
It seems to be easy I tried to prove it.I first tried some cases.
$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?
update1: the numerator should be prime

How to solve: $\underset{n\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{n^n+\frac{1}{n}}{(n+\frac{1}{n}{)}^n}{t}^n$
t is a whole number.
Thank you very much! Please tell me your ideas, even if you won't post the the result.

Solve quadratic fraction
I would like to simplify the fraction
$\frac{x^2-<!--\; -\; -->2x}{x^2+x-<!--\; -\; -->6}$
I know from Mathematica that it should equal $\frac{3}{3x}$ but how do I get there?

One tap fills a pool. The other one empties it. It's a word problem.
In a pool there are two taps, one for filling and one for emptying. Once, when the pool was empty they opened the filling tap for 4 hours. Afterwards, they opened by mistake the emptying tap and after 2 hours the pool was filled 80% from its volume. When the pool was completely filled it turned out that the filling tap filled like water amount of $1\frac{1}{2}$ pools and the emptying tap emptied water amount of $\frac{1}{2}$ pool.
So what I did is
$\overline{)\begin{array}{cccc}& \text{power / rate}& \text{time (hours)}& \text{Total}\\ \text{Filling tap until 80\%}& \frac{1}{x}& 6& \frac{1}{x}\times <!--\; \times \; -->6=\frac{6}{x}\\ \text{Emptying tap until 80\%}& \frac{1}{y}& 2& \frac{1}{y}\times <!--\; \times \; -->2=\frac{2}{y}\\ \text{Filling tap until the end}& \frac{1}{x}& \frac{1.5}{x}& 1.5\\ \text{Emptying tap until the end}& \frac{1}{y}& \frac{0.5}{y}& 0.5\end{array}}$
I can write that:
$\frac{6}{x}-<!--\; -\; -->\frac{2}{y}=0.8$
And here I'm stuck. I don't know how to use the information of the last two rows of my chart. If I would know the time it took to fill 100% the pool then I could sum up the two last rows from the middle column. Or the time it took to fill up the remaining 20%.
Sorry if my translation of the question wasn't so good.

Struggling to reach answer with algebraic fractions
The question is
$\frac{2x^3-<!--\; -\; -->x^{-<!--\; -\; -->3}}{x^3}$
and the answer is
$2-<!--\; -\; -->x^{-<!--\; -\; -->6}$
I have been trying but cannot get to the correct answer using the methods I know. If anyone can explain to me how to do it I would be extremely grateful.

Fraction Transforms
Here's a number theory problem I'm having some difficulty with:
Say we transform a fraction by the following rule: we start with some fraction $\frac{m}{n}$ with $m>n$ and then convert it to $\frac{fr}{n}$, where $f=\lfloor <!--\; \lfloor \; -->m/n\rfloor <!--\; \rfloor \; -->$ and $r=mmodn$. We repeat the process until our fraction is less than 1 or an integer. Given an n is there an m so that we end up with a fraction less than 1?
For $n=2$ it isn't that hard, 3, 7, 15, 31, etc. work, but what about other values of $n$?