Explore Fraction Concepts with Examples and Questions

Divide or state that the division is undefined: (Note: If applicable, type your answer as a fraction, not as a decimal.)
A) -2/5 divide (- 3/4)
B) 3/0

Divisor of fraction is divisor of numerator or denominator?
If we have
$q\mid <!--\; \mid \; -->\frac{x^a–1}{x–1},$
does that mean that $q$ is divisor of $x^a–1$ or $q$ is divisor of $x-<!--\; -\; -->1$?

For $a,b,c<1$. Minimize $M=\frac{a^2(1-<!--\; -\; -->2b)}{b}+\frac{b^2(1-<!--\; -\; -->2c)}{c}+\frac{c^2(1-<!--\; -\; -->2a)}{a}$
My Try: From $ab+bc+ca=1\iff <!--\; \iff \; -->\sqrt{3}\le <!--\; \le \; -->a+b+c<3$
We have: $M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-<!--\; -\; -->2(a^2+b^2+c^2)$
$\ge <!--\; \ge \; -->\frac{{(a+b+c)}^2}{a+b+c}-<!--\; -\; -->2(a^2+b^2+c^2)-<!--\; -\; -->4(ab+bc+ca)+4(ab+bc+ca)$
$=a+b+c-<!--\; -\; -->2(a+b+c{)}^2+4(1)$
Let $a+b+c=x(\sqrt{3}\le <!--\; \le \; -->x<3)$
We find Min of function $y=-<!--\; -\; -->2x^2+x+4$
And $y_{Min}=-<!--\; -\; -->2+\sqrt{3}$ when $x=\sqrt{3}$
Right or Wrong ?

Equation in a fraction - have I solved it right?
I'm solving entrance maths example problems from a university. I solved this one:
$\frac{2x-<!--\; -\; -->3}{5}-<!--\; -\; -->\frac{4x+5}{3}=8$
My solution was -104/14 but in the answer sheet it's -11. Am I wrong? If so why?
Edit: I did the following steps:
1.$3(2x-<!--\; -\; -->3)-<!--\; -\; -->5(4x+5)=8$
2.$6x-<!--\; -\; -->9-<!--\; -\; -->20x+25=120$
3.$-<!--\; -\; -->14x=104$
4.$x=-<!--\; -\; -->104/14$

Algorithm for converting fraction into (recurring) decimal
I need to determine whether fraction is recurring decimal or not (what are conditions for it?), find period and output it as $\frac{1}{3}=0.\stackrel{¯<!--\; ¯\; -->}{3}$
If it is not recurring, then I already have algorithm.

$f(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})+f(\frac{4\mathrm{\pi <!--\; \pi \; -->}}{7})+f(\frac{6\mathrm{\pi <!--\; \pi \; -->}}{7})=1$
let
$f(x)=\frac{1}{1+2\cos <!--\; \; -->x}$
prove that :
$f(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})+f(\frac{4\mathrm{\pi <!--\; \pi \; -->}}{7})+f(\frac{6\mathrm{\pi <!--\; \pi \; -->}}{7})=1$
My Try :
$f(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})=\frac{1}{1+2\cos <!--\; \; -->(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})}$
$f(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})=\frac{1}{1+2\cos <!--\; \; -->(\frac{4\mathrm{\pi <!--\; \pi \; -->}}{7})}$
$f(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})=\frac{1}{1+2\cos <!--\; \; -->(\frac{6\mathrm{\pi <!--\; \pi \; -->}}{7})}$
$L=\frac{1}{1+2\cos <!--\; \; -->(\frac{6\mathrm{\pi <!--\; \pi \; -->}}{7})}+\frac{1}{1+2\cos <!--\; \; -->(\frac{4\mathrm{\pi <!--\; \pi \; -->}}{7})}+\frac{1}{1+2\cos <!--\; \; -->(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{7})}$
what now ?

Zero/Zero questions and perhaps faulty logic
So I only have an Algebra II level understanding of math seeing as I am still in high school and am still missing some fundamentals seeing as I didn't pay attention in math until this year. However when recalling something my algebra teacher had taught me during the year I came up with some questions regarding the logic recently.
So during the school year, I was taught that $\frac{2}{2}=1,\frac{a}{a}=1,\frac{xy}{xy}=1$ and so forth but $\frac{0}{0}=\text{Undefined}$... and while researching this topic I found that the algebraic way to write all these fractions is as such $2(x)=2,a(x)=a,$ and upon researching this further I found that the reason that $\frac{0}{0}$ is undefined is that for any value of $x$ the equation holds true. However, seeing as in the fraction $\frac{a}{a}$ $a$ is a variable and variables can represent any given quantity I was wondering in the case that $a=0$ would $\frac{a}{a}$ still $=1$ and if not why along with the fact that lets say $a=0$ and you didn't know it why is it safe to assume that $a$ would never equal zero? Also if it happens to be the case where when $a=0,\frac{a}{a}=1$ (which I doubt it is) shouldn't this mean that $\frac{0}{0}=1$ then?

Partial Fractions help!?
$A+C=0$
$-<!--\; -\; -->4A+B-<!--\; -\; -->8C+D=1$
$3A+16C-<!--\; -\; -->8D=-<!--\; -\; -->29$
$-<!--\; -\; -->12A+3B+16D=5$
How do I equate the coefficients? Please provide steps an an explanation.

Solve $2^x\cdot <!--\; \cdot \; -->6^{x-<!--\; -\; -->2}=5^{2x}\cdot <!--\; \cdot \; -->7^{1-<!--\; -\; -->x}$
I have to solve this equation, $2^x\cdot <!--\; \cdot \; -->6^{x-<!--\; -\; -->2}=5^{2x}\cdot <!--\; \cdot \; -->7^{1-<!--\; -\; -->x}$. Now, I started by taking logs on both sides which gives me this funny looking equation
$x\log <!--\; \; -->2+(x-<!--\; -\; -->2)\log <!--\; \; -->(2\cdot <!--\; \cdot \; -->3)=2x\log <!--\; \; -->(\frac{10}{2})+(1-<!--\; -\; -->x)\log <!--\; \; -->7$
I have been stuck on this step for a while now and can't see how I can go further from here. Is there a way out?

Solving $Т(n)=\frac{1}{\frac{1}{T(n-<!--\; -\; -->1)}+n^2}$ and $T(1)=1$
I did find $T(2)=\frac{1}{1+n^2}$, but I don't know how to proceed.
How do I go on from here? How can I find the solution? Thanks. :)

Change $0.75$ to a fraction and reduce to lowest terms

partial fraction decomposition of $\frac{k^4}{(a{k}^3-<!--\; -\; -->1{)}^2}$
I have to perform complex partial fraction decomposition of the following term:
$\frac{k^4}{(a{k}^3-<!--\; -\; -->1{)}^2}$
where $a$ is a real positive number.
and I would like to know if it is possible to reduce it to a sum of fractions of the type $\frac{A}{k\pm <!--\; \pm \; -->z}$,$\frac{Bk}{k^2\pm <!--\; \pm \; -->z}$,$\frac{Ck}{k^2-<!--\; -\; -->y}$ or similar. Where $z$ is a complex number and $y$ is a real number.
If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too.
Any hint on the process, or any reference would be nice.
Thanks in advance

Weird large K symbol
Take a look at this symbol:
$\mathrm{\pi <!--\; \pi \; -->}=3+\underset{k=1}{\stackrel{\mathrm{\infty <!--\; \infty \; -->}}{\mathrm{K}}}\frac{(2k-<!--\; -\; -->1{)}^2}{6}$
Does it look familiar to you? If so please help me!

Can a negative number be raised to a fractional power e.g $2.5$?
Is it possible to raise a negative integer or other negative number to a power that is not an integer ?
I am asking because when plotting some polynomial functions the negative numbers were not plotted and I am curious why.
Why is it possible to raise $-<!--\; -\; -->5$ to the power of $7$ but it is not possible to raise $-<!--\; -\; -->5$ to the number of $7.5$ and is it also impossible to plot the function $x^{7.5}$?

Fraction inequality
Suppose we know that
$\begin{array}{r}\frac{a+b+d}{a^{\prime }+b^{\prime }+d^{\prime }}\le <!--\; \le \; -->M\text{ and }\frac{a+c+d}{a^{\prime }+c^{\prime }+d^{\prime }}\le <!--\; \le \; -->M\end{array}$
and
$\begin{array}{r}\frac{a}{a^{\prime }}\le <!--\; \le \; -->\frac{b}{b^{\prime }}\le <!--\; \le \; -->\frac{d}{d^{\prime }}\text{ and }\frac{a}{a^{\prime }}\le <!--\; \le \; -->\frac{c}{c^{\prime }}\le <!--\; \le \; -->\frac{d}{d^{\prime }}\end{array}$
and also that $a,b,c,d,a^{\prime },b^{\prime },c^{\prime },d^{\prime }\in <!--\; \in \; -->(0,1]$
Is the following true?
$\begin{array}{r}\frac{a+b+c+d}{a^{\prime }+b^{\prime }+c^{\prime }+d^{\prime }}\le <!--\; \le \; -->M\end{array}$

Prove equivalent fractions with $\pm <!--\; \pm \; -->$ sign
If
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{n_1}{n_2}=\frac{m_1}{m_2}$
prove that
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{n_1\pm <!--\; \pm \; -->m_1}{n_2\pm <!--\; \pm \; -->m_2}$
I know this is true if I add numbers to it
$\frac{1}{2}=\frac{2}{4}$
$\frac{1+2}{2+4}=\frac{3}{6}=\frac{1}{2}$
$\frac{1-<!--\; -\; -->2}{2-<!--\; -\; -->4}=\frac{-<!--\; -\; -->1}{-<!--\; -\; -->2}=\frac{1}{2}$

proving that $\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{d}}=\frac{3}{\sqrt{a}+\sqrt{d}}$ for any A.P.
if $a,b,c,d$ are an arithmetic progression (in that order), prove that
I made $n$ the common difference of $a,b,c,d$; so
$a=a$
$b=a+n$
$c=a+2n$
$d=a+3n$
I tried to replace the terms with those, anyways i squared both equalities but i didn 't get nothing since i'm pretty bad with square roots. I'm looking for some hints or properties that can be useful. Thanks

Showing $\frac{1}{2}{3}^{2-<!--\; -\; -->p}(x+y+z{)}^{p-<!--\; -\; -->1}\le <!--\; \le \; -->\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking $p_1=p_2=p_3=\frac{1}{3}$, (say $x=x_1,y=x_2,z=x_3$)
${(\sum <!--\; \sum \; -->p_k{x}_k)}^p\le <!--\; \le \; -->\sum <!--\; \sum \; -->p_k{x}^p$ which means;
$3^{1-<!--\; -\; -->p}(x+y+z{)}^p\le <!--\; \le \; -->x^p+y^p+z^p$
and if I assume wlog $x\ge <!--\; \ge \; -->y\ge <!--\; \ge \; -->z$ then $\frac{1}{y+z}\ge <!--\; \ge \; -->\frac{1}{x+z}\ge <!--\; \ge \; -->\frac{1}{x+y}$
so these $2$ sequences are similarly ordered, and by rearrangement inequality the rest follows ?
I should use power means, but any other solution is also appreciated.

Prove that $\frac{(2a+b+c{)}^2}{2a^2+(b+c{)}^2}+\frac{(2b+c+a{)}^2}{2b^2+(c+a{)}^2}+\frac{(2c+a+b{)}^2}{2c^2+(a+b{)}^2}\le <!--\; \le \; -->8$
Prove that
$\frac{(2a+b+c{)}^2}{2a^2+(b+c{)}^2}+\frac{(2b+c+a{)}^2}{2b^2+(c+a{)}^2}+\frac{(2c+a+b)}{2c^2+(a+b{)}^2}\le <!--\; \le \; -->8$
MY ATTEMPT:I want to make a relation between $a,b,c$. By trial I found that if we put $a=b=c=1$ then the above inequality holds(equality also holds). So by trial I assume that $a+b+c=3$. After that the three functions become of the form of the function below: $f(x)=\frac{(x+3{)}^2}{2x^2+(3-<!--\; -\; -->x{)}^2}$. I calculate the function and found that : $f(x)\le <!--\; \le \; -->⅓(4x+4)$. Am I do right . Anybody has other ideas.

Is $9\ast <!--\; \ast \; -->(1/9)=\mathrm{0.99999999...}$ statement correct?
We can tell that
$\mathrm{0.11111111...}\cdot <!--\; \cdot \; -->9=\mathrm{0.999999999...}$
And that
$\frac{1}{9}=\mathrm{0.11111111111...}$
Therefore
$\frac{1}{9}\cdot <!--\; \cdot \; -->9=\mathrm{0.999999999...}$
However, we know that
$\frac{1}{9}\cdot <!--\; \cdot \; -->9=\frac{9}{9}=1$
Note: I'm taking in account that ... are the other rational digits left.
What am I making wrong? What is misunderstood?
Thanks for the help in clearing this problem.