Explore Fraction Concepts with Examples and Questions

I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired.
$\{\begin{array}{l}\frac{c_1}{1-<!--\; -\; -->x_1}+\frac{c_2}{1-<!--\; -\; -->x_2}+\frac{c_3}{1-<!--\; -\; -->x_3}=0\\ \frac{c_1}{1-<!--\; -\; -->x_4}+\frac{c_2}{1-<!--\; -\; -->x_5}+\frac{c_3}{1-<!--\; -\; -->x_6}=0\\ c_1\ln <!--\; \; -->\frac{x_1}{1-<!--\; -\; -->x_1}+c_2\ln <!--\; \; -->\frac{x_2}{1-<!--\; -\; -->x_2}+c_3\ln <!--\; \; -->\frac{x_3}{1-<!--\; -\; -->x_3}=0\\ c_1\ln <!--\; \; -->\frac{x_4}{1-<!--\; -\; -->x_4}+c_2\ln <!--\; \; -->\frac{x_5}{1-<!--\; -\; -->x_5}+c_3\ln <!--\; \; -->\frac{x_6}{1-<!--\; -\; -->x_6}=0\\ \frac{x_1(1-<!--\; -\; -->x_1)}{x_4(1-<!--\; -\; -->x_4)}=\frac{x_2(1-<!--\; -\; -->x_2)}{x_5(1-<!--\; -\; -->x_5)}=\frac{x_3(1-<!--\; -\; -->x_3)}{x_6(1-<!--\; -\; -->x_6)}>1\end{array}$
where $c_1,c_2,c_3$ are constants, and $x_i\in <!--\; \in \; -->(0,1),i=1,2,3,4,5,6$

Simplification of rational expressions
I have the following expression:
$\frac{2}{x-<!--\; -\; -->2}+\frac{2}{x^2-<!--\; -\; -->5x+6}$
So I can simplify this as:
$\frac{2}{x-<!--\; -\; -->2}+\frac{2}{(x-<!--\; -\; -->3)(x-<!--\; -\; -->2)}$
I make the common denominator to be $(x-<!--\; -\; -->3)(x-<!--\; -\; -->2)$
So I then apply $(x-<!--\; -\; -->3)$ to the left hand side which gives me:
$\frac{2(x-<!--\; -\; -->3)+2}{(x-<!--\; -\; -->3)(x-<!--\; -\; -->2)}$
I have clearly taken a wrong step because the answer in the book to the original expression is $\frac{2}{x-<!--\; -\; -->3}$ so I'm not sure how that answer was arrived at.

Rationalization of a fraction
Can someone please explain how i can rationalize the fraction $\frac{2+\sqrt{3}}{(2-<!--\; -\; -->\sqrt{3}{)}^3}$ so that i obtain an answer that has an exponent of 6. I basically need to compare this value with ${\displaystyle \frac{(10{)}^6}{(3{)}^6}}$ to see which is greater. Thanks in advance for helping out!

Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
If $\frac{x}{y}$ + $\frac{y}{x}$ = 3

How to solve following limit
I've been struggeling a bit with the following limit:
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{a-<!--\; -\; -->\sqrt{a^2-<!--\; -\; -->x^2}}{x^2}$
The solution is:
If a < 0 then $\mathrm{\infty <!--\; \infty \; -->}$. If a > 0 then $\frac{1}{2a}$
But I don't know how to get there.
Thank you

Writing the sum of two rational functions as a single rational function.
Write as a single fraction:
$\frac{2x}{x-<!--\; -\; -->1}-<!--\; -\; -->\frac{x}{x+1}$
Please can somebody talk me through this question as I don't understand how to get a common denominator. Thank you.

Primes and certain unit fractions
Are there primes $p,q$ and a natural number $a$ such that $\frac{1}{p}+\frac{1}{q}=\frac{1}{a}$?

How does $-<!--\; -\; -->\frac{1}{x-<!--\; -\; -->2}+\frac{1}{x-<!--\; -\; -->3}$ become $\frac{1}{2-<!--\; -\; -->x}-<!--\; -\; -->\frac{1}{3-<!--\; -\; -->x}$
I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-<!--\; -\; -->\frac{1}{x-<!--\; -\; -->2}+\frac{1}{x-<!--\; -\; -->3}$ becomes $\frac{1}{2-<!--\; -\; -->x}-<!--\; -\; -->\frac{1}{3-<!--\; -\; -->x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$\begin{array}{rl}\frac{1}{x^2-<!--\; -\; -->5x+6}& =\frac{1}{(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)}=\frac{1}{-<!--\; -\; -->3-<!--\; -\; -->(-<!--\; -\; -->2)}(\frac{1}{x-<!--\; -\; -->2}-<!--\; -\; -->\frac{1}{x-<!--\; -\; -->3})=-<!--\; -\; -->\frac{1}{x-<!--\; -\; -->2}+\frac{1}{x-<!--\; -\; -->3}\\ & =\frac{1}{2-<!--\; -\; -->x}-<!--\; -\; -->\frac{1}{3-<!--\; -\; -->x}=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{2^{n+1}}{x}^n-<!--\; -\; -->\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{3^{n+1}}{x}^n=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\frac{1}{2^{n+1}}-<!--\; -\; -->\frac{1}{3^{n+1}})x^n\end{array}$

Triple fractions (and more complex fractions)
Usually
$\frac{a}{\frac{b}{c}}=\frac{ac}{b}$
i.e. $b/c$ is seen as the denominator, and $a$ is the numerator.
If you have $a/b/c/d$, what do you choose to take as the denominator?
$\frac{b}{\frac{c}{d}},\frac{c}{d},\text{ or }d\text{?}$
and why?
And what about for $a/b/c/d/e$?

Simple Fraction needing explanation
$\frac{x}{x^{-<!--\; -\; -->1/2}}=x^{3/2}$
How? I don't see what is going on here. What rule is being used to achieve this amount?

Prove the inequalitiy
If $a,b,c$ are positive reals such that: $\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{b+a+1}\le <!--\; \le \; -->1$ then:
$\frac{1}{b+c+1}+\frac{1}{a+c+1}+\frac{1}{b+a+1}\ge <!--\; \ge \; -->1$
I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.

Prove that one of the elements can't be in the interval $(0,1)$
Let be $a,b,c\in <!--\; \in \; -->\mathbb{R}$ so that the sum of two of them is never equal to $1$. Prove that atleast on of $\frac{ab}{a+b-<!--\; -\; -->1},\frac{bc}{b+c-<!--\; -\; -->1},\frac{ca}{c+a-<!--\; -\; -->1}$ can't be in the interval $(0,1)$
I aproached it with contradiction, but can't get sth that is not true.

Please help me understand this simple fraction
I got started learning about fractions a few days ago, The tutorial I'm using for this, is limited to fractions like this
$g)\frac{7}{12}\times <!--\; \times \; -->\frac{8}{21}=\frac{2}{9}$
Now as I'm trying to find further exercices, I keep stumbling upon fraction-based problems that are built like this one.
5.Myltiply.
$a)5\frac{1}{4}\times <!--\; \times \; -->2\frac{2}{3}=14$
What does the number on the left mean? I mean the huge 5 and huge 2. is it a whole + a fraction? like 5 + (1/4)??
I would like to further my research on this topic, but I don't know what the name of these types of fractions are.
receiving a link to a tutorial would be fine too, I just don't know what search terms to use.
Also why is the solution 14?

Partial fractions (How do I get from x to y)
how do I get from
$\mathrm{\gamma <!--\; \gamma \; -->}\ast <!--\; \ast \; -->\left(\frac{\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}_{0}w}{(1+r{)}^t{\mathrm{\beta <!--\; \beta \; -->}}^t\mathrm{\alpha <!--\; \alpha \; -->}}}{\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}_{0}w}{(1+r{)}^t{\mathrm{\beta <!--\; \beta \; -->}}^t\mathrm{\alpha <!--\; \alpha \; -->}}-<!--\; -\; -->\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}_2}{{\mathrm{\beta <!--\; \beta \; -->}}^t\mathrm{\alpha <!--\; \alpha \; -->}}}\right)$
to
$\frac{\mathrm{\gamma <!--\; \gamma \; -->}rw{\mathrm{\lambda <!--\; \lambda \; -->}}_0{\mathrm{\beta <!--\; \beta \; -->}}^t}{w{\mathrm{\lambda <!--\; \lambda \; -->}}_0{\mathrm{\beta <!--\; \beta \; -->}}^t-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_2(\mathrm{\beta <!--\; \beta \; -->}(r+1){)}^t}$
Im stuck. Any help?

When the numerator of a fraction is increased by 4, the fraction increases by 2/3...
When the numerator of a fraction is increased by 4, the fraction increases by 2/3. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be x and the denominator be y.
Accordingly,
$\frac{x+4}{y}=\frac{x}{y}+\frac{2}{3}$
I am not able to find the second equation.

How can I find x, y values for $\frac{(1+i)x-<!--\; -\; -->2i}{3+i}+\frac{(2-<!--\; -\; -->3i)y+i}{3-<!--\; -\; -->i}=i$
I believe the format I need in order to solve this problem should be such that the real parts and imaginary parts are separated, $\{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\}+\{\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\}i=i$
Then I can equate the real and imaginary parts of the equation and solve for x and y. I tried to multiply by the conjugate so that I would get a real number at the denominator.
Rearranged:
$\frac{x+i(x-<!--\; -\; -->2)}{3+i}+\frac{2y+i(-<!--\; -\; -->3y+1)}{3-<!--\; -\; -->i}=i$
Multiplied by conjugate:
$\frac{x(3-<!--\; -\; -->i)+i(x-<!--\; -\; -->2)(3-<!--\; -\; -->i)}{10}+\frac{2y(3+i)+i(-<!--\; -\; -->3y+1)(3+i)}{10}=i$
And at this step I started to feel as if I made a mistake because I weren't sure how to proceed. Would someone let me know if my approach was correct and show me how to do this?

Show that for every integer $1+\frac{1}{4}+\frac{1}{9}+···+\frac{1}{n^2}\le 2-\frac{1}{n}$

How did he get the fraction with fraction power?
So we have a simple equation that is from Kepler.
${\left(\frac{{\stackrel{¯<!--\; ¯\; -->}{r}}_1}{{\stackrel{¯<!--\; ¯\; -->}{r}}_2}\right)}^3={\left(\frac{T_1}{T_2}\right)}^2$
In an explanation of a physics book, you can resolve for $r_2$ like this:
$r_2=r_1{\left(\frac{T_1}{T_2}\right)}^{2/3}$
And I found
$r_1=\sqrt[3]{\frac{T_1^2}{T_2^2}{r}_2^3}$
First question, is my approach correct? My second and main question is how did he get the $r_2$ equation that I stated first. The physics book doesn't explain how to get from the main equation to $r_2=r_1{\left(\frac{T_1}{T_2}\right)}^{2/3}$. Can someone explain me, please? (By the way, of course the equation for $r_1$ and $r_2$ should be different).
Thank you!

Prove $\sum <!--\; \sum \; -->\frac{1}{(x+2y{)}^2}\ge <!--\; \ge \; -->\frac{1}{xy+yz+zx}$
Let $x,y,z>0$ Prove that:
$\frac{1}{(x+2y{)}^2}+\frac{1}{(y+2z{)}^2}+\frac{1}{(z+2x{)}^2}\ge <!--\; \ge \; -->\frac{1}{xy+yz+zx}$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!

another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le <!--\; \le \; -->\frac{1}{2}$
Let $x,y,z>0$ and such $xyz\ge <!--\; \ge \; -->2+x+y+z$, show that
$\begin{array}{}\text{(1)}& {\displaystyle \frac{1}{xy+z}}+{\displaystyle \frac{1}{yz+x}}+{\displaystyle \frac{1}{zx+y}}\le <!--\; \le \; -->{\displaystyle \frac{1}{2}}\end{array}$
The theory basis of speculation Use the following classical results
$xyz=2+x+y+z⟹<!--\; ⟹\; -->xy+yz+zx\ge <!--\; \ge \; -->2(x+y+z)$