Explore Fraction Concepts with Examples and Questions

Fraction simplification Rules
I am studying for GRE and One of the practice questions is a division. After converting my Mixed numeral I get $\frac{90}{72}$ now I just have to simplify.
What I understood is that you divide by Least common multiplier so I
1.) Divide $\frac{90}{72}$ by 2: And I got $\frac{45}{36}$
2.) Divide $\frac{45}{36}$ by 3: And I got $\frac{9}{12}$
3.) and so on until I got to 3/4 which is the wrong answer
If in step 1, I would have divide by 3
1.) Divide $\frac{90}{72}$ by 3 and I got $\frac{10}{8}$ 2.) Divide $\frac{10}{8}$by 2 and I got \frac{5}{4} which is the right answer
What is the rule to simplify fractions?

Proof of an inequality involving different exponents
I was asked the following interesting question but I can not come up with a proof.
Let $a,b,c$ be positive real numbers. Prove that
$\begin{array}{}\text{(}\mathrm{♣<!--\; ♣\; -->}\text{)}& \frac{a^3}{b^3+c^3}+\frac{b^3}{a^3+c^3}+\frac{c^3}{a^3+b^3}\ge <!--\; \ge \; -->\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\end{array}$
One way to prove $(\mathrm{♣<!--\; ♣\; -->})$ may be to consider the following function
$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{a^x+b^x}$
and prove that $f(x)$ is a non-decreasing function on $[2,3]$. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?

Division by rational (decimal) number meaning
When I say, that I exchanged 42 CZK into 1,5 euro. Why do I get the rate for one euro by dividing?
1) How do you explain this division in words. Like when you say when doing integer division, that you divide 10 cookies to 2 people, so one get 5. That type of style.
2) Does division always gives back an average rate per 1 unit of somethings?
3) the same for mean average, I bought 100 units of something for 2500 unit of any currency. 2500/100 = 25 per one (25/1).
How do you explain this in words too, please. Something like you want to distribute 2500 into 100 equal parts and the answer is the units per one?
Thanks, All I care is the problems being said in words, meaning of division rational number by rational number, in this case in its decimal form

Let $P$ be a polynomial of degree $n$ , $n\ge <!--\; \ge \; -->2$, $P(X)=\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}{(X-<!--\; -\; -->z_i)}^{m_i}$
with zeros $z_1,z_2,\dots <!--\; \dots \; -->,z_n$
Decompose the rational fraction ${\displaystyle \frac{P^{\prime }}{P}}$
My thoughts:
$P(X)=\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}{(X-<!--\; -\; -->z_i)}^{m_i}$
$P^{\prime }(X)=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{m}_i{(X-<!--\; -\; -->z_i)}^{m_i-<!--\; -\; -->1}\underset{?}{\overset{?}{\prod <!--\; \prod \; -->}}{(X-<!--\; -\; -->z_i)}^{m_{?}}$
I don't know with what i should fill the ?
Then:
$\begin{array}{rl}{\displaystyle \frac{P^{\prime }}{P}}& ={\displaystyle \frac{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{m}_i{(X-<!--\; -\; -->z_i)}^{m_i-<!--\; -\; -->1}\underset{?}{\overset{?}{\prod <!--\; \prod \; -->}}{(X-<!--\; -\; -->z_i)}^{m_i}}{\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}{(X-<!--\; -\; -->z_i)}^{m_i}}}\\ & =?\end{array}$
I'm stuck here

Trivial Summation Challenge
Write the repeating decimal $\mathrm{1.367367367367...}$ in the form $a/b$ where both $a$ an $b$ are positive whole numbers, but do this using a converging infinite sum.

Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le <!--\; \le \; -->\frac{3}{2}$
For positive real numbers with $a+b+c=abc$ prove that
$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le <!--\; \le \; -->\frac{3}{2}$
I made the substitution $a=\tan <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->}),b=\tan <!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->}),c=\tan <!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})$ with the constraint $\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->}+\mathrm{\gamma <!--\; \gamma \; -->}=\mathrm{\pi <!--\; \pi \; -->}$
My inequality reduces to proving,
${\cos }^2<!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})+{\cos }^2<!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})+{\cos }^2<!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})\le <!--\; \le \; -->\frac{3}{2}$
But I am stuck on it. Any help with either inequality would be appreciated.

Can someone explain why this happens? (Dividing variables with exponents)
Alright, so let's say I have
$\frac{x^{-<!--\; -\; -->6}}{-<!--\; -\; -->x^{-<!--\; -\; -->4}}$
The answer is ${\displaystyle \frac{1}{x^2}}$, but why isn't it ${\displaystyle \frac{1}{-<!--\; -\; -->x^2}}$??

Why does ${\displaystyle \frac{8}{\frac{8\sqrt{145}}{145}}}=\sqrt{145}$?

Fast way to calculate Fraction
I want to solve this problem not by using common denominator. Is there any innovative and fast way ?.
$\frac{1}{6}+\frac{4}{21}+\frac{5}{84}+\frac{7}{228}+\frac{9}{532}=?$

Basic Algebra combining exponent fractions/simplifying (George F Simmons "Precalculus in Nutshell")
From George F. Simmons 'Precalculus' book, Algebra section, 5(d); Combine and Simplify
$\frac{x}{x{y}^2}+\frac{y}{x^{2}y}$
Combine: =
$\frac{x(x^{2}y)+y(x{y}^2)}{(x{y}^2)(x^{2}y)}$
Simplify: =
$\frac{x(x^{2}y)+y(x{y}^2)}{xy(x^2{y}^2)}$
The given answer, which I can't figure out how he arrives at it is:
$\frac{x^2+y^2}{x^2{y}^2}$

Fraction confusion
I read in a set of memoranda that if $\frac{b-<!--\; -\; -->x}{x}=\frac{b}{a}$, then
$x=\frac{ab}{a+b}$
How is this true? I tried working it out but I could not understand. Please help.

Help with Calculate integral
Find ${\int <!--\; \int \; -->}_0^a{\displaystyle \frac{3x^2-<!--\; -\; -->ax}{(x-<!--\; -\; -->2a)(x^2+a^2)}}dx$ I tried using partial fractions and the substitution $u=a-<!--\; -\; -->x$ but I haven't made any real progress. Please help.

Partial fraction in two variable problem
How to write partial fraction of
$\frac{12m-<!--\; -\; -->n-<!--\; -\; -->3mn+7}{5m-<!--\; -\; -->2n-<!--\; -\; -->2mn+5}$
I just write first and second denominator: $5-<!--\; -\; -->2n$ and $m+1$

Showing this fraction is less than one?
And also, what are some techniques to show such a thing. That is, what are standard techniques for showing a fraction is less than one? I'm just trying to refresh/improve/add to my toolbox.
The fraction I am working with is:
$\frac{tx(tx+1)+t^2}{(tx+1{)}^2}$
The problem is: Show
$\frac{tx(tx+1)+t^2}{(tx+1{)}^2}\le <!--\; \le \; -->1\text{ for }x\ge <!--\; \ge \; -->t,t\ge <!--\; \ge \; -->0$
I have been trying to move the denominator to the right and show that the numerator is less than the denominator, but I am not seeing any clean cancellations, or obvious result.

Show that $\frac{1}{a}+\frac{1}{b}\ne \frac{1}{a+b}$
Problem
Assume that $a,b\in <!--\; \in \; -->\mathbb{R}-<!--\; -\; -->\{0\}$ and that $a+b\ne 0$. Prove that $\frac{1}{a}+\frac{1}{b}\ne \frac{1}{a+b}$
My Proof
Let's assume that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, then it follows that
$(a+b{)}^2-<!--\; -\; -->ab=0$
Let $x=a+b$ and $y=ab$. Now $b=x-<!--\; -\; -->a$ and so $y=a(x-<!--\; -\; -->a)=ax-<!--\; -\; -->a^2$. The previous equation can be written as
$x^2-<!--\; -\; -->y=0$
Substituting $y=ax-<!--\; -\; -->a^2$ in this equation gives
$x^2-<!--\; -\; -->ax+a^2=0$
The discriminant of this quadratic equation (in x) is $-<!--\; -\; -->3a^2<0$ and therefore $x=a+b$ has no real solution. This means $a+b\in <!--\; \in \; -->\mathbb{C}$ and therefore either of a or b or both are not real but this contradicts our assumption that a,b are real numbers. Therefore by contradiction $\frac{1}{a}+\frac{1}{b}\ne \frac{1}{a+b}$
My Question
Is my proof correct? are there any alternative proofs?

Why Doesn't $2^{1/n}=1/(2^n)$
Take $2^{1/n}$. Since $1/n$ can be simplified as $n^{-<!--\; -\; -->1}$, the original term can become $2^{n^{-<!--\; -\; -->1}}$. The exponents can then be multiplied to result in $2^{-<!--\; -\; -->n}$ which is $1/(2^n)$. However it is obvious that
$2^{1/n}\ne <!--\; \ne \; -->1/(2^n)$
because the first item approaches 1 as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ but the second item approaches 0. What is wrong with the logic I used to simplify the first into the second that causes this discrepancy?

How to solve the following equation: $\frac{3}{x}-<!--\; -\; -->\frac{x-<!--\; -\; -->3}{2x+10}+\frac{8}{3x+15}=\frac{4}{3}$?

The result of subtracting the integer part of $x$ from $x$
I want to know what does this formula do with the integer $x$
$\text{frac}(x)=x-<!--\; -\; -->\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->,x>=0$
I've searched and found that this is called finding the fraction part of $x$
Can anyone explain it for me?

Expressing $\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ with rational Denominator

Question in fraction (not simple )
I have a question and its answer but I don't know how can i solve
$\frac{37}{13}=2+\frac{1}{x+\frac{1}{5+\frac{1}{y}}}$
the answer $x=1,y=2$
Could any one explain how to solve this ?? please