Explore Fraction Concepts with Examples and Questions

Find:
$$\frac{4}{y}-<!--\; -\; -->\frac{5}{3}=-<!--\; -\; -->\frac{1}{5}$$

How to show the following is true?
I need a proof of the following. If
$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots <!--\; \cdots \; -->=\frac{a_K}{b_K}=T$$
then
$$\frac{a_i}{b_i}=\frac{\underset{i=1}{\overset{K}{\sum <!--\; \sum \; -->}}{a}_i}{\underset{i=1}{\overset{K}{\sum <!--\; \sum \; -->}}{b}_k}$$
where all the $a_i$'s and $b_i$'s are positive values.

In resticted domain , Applying the Cauchy-Schwarz's inequality
I see this problem a few days ago.
$$a,b,c\in <!--\; \in \; -->[\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}]$$
prove that
$$9\le <!--\; \le \; -->(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le <!--\; \le \; -->\frac{(2\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->})(\mathrm{\alpha <!--\; \alpha \; -->}+2\mathrm{\beta <!--\; \beta \; -->})}{\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{\beta <!--\; \beta \; -->}}$$
Left inequality is easy by Cauchy-Schwart's inequality. But Right inequality is difficult to me. How can I approach the Right inequlity?
EDIT : Sorry, I forgot the condition : $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0$

What is the sum of all integers $n$ such that $n^2+2n+2$ divides $n^3+4n^2+4n-<!--\; -\; -->14$?
How to proceed in this problem? I divided $n^3+4n^2+4n-<!--\; -\; -->14$ by $n^2+2n+2$ and equated the remainder to zero . I got $n=5$ as answer but the answer is wrong.

Prove that $\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\le <!--\; \le \; -->\frac{3}{2}$
Let $abc=1$ and $a,b,c>0$. Prove that
$$\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\le <!--\; \le \; -->\frac{3}{2}.$$
I guess that
$$\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}=\frac{3}{2}$$
if and only if $a=b=c=1$. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.

Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge <!--\; \ge \; -->\frac{3}{2}$
For $a\ge <!--\; \ge \; -->b\ge <!--\; \ge \; -->c>0$. Prove that
$a=100;b=1;c=1/100$ it's wrong ???
${\displaystyle \frac{a}{a+b}}+{\displaystyle \frac{b}{b+c}}+{\displaystyle \frac{c}{c+a}}\ge <!--\; \ge \; -->{\displaystyle \frac{3}{2}}$
$<=>\sum <!--\; \sum \; -->\frac{a}{a+b}-<!--\; -\; -->\frac{1}{2}+\frac{b}{b+c}-<!--\; -\; -->\frac{1}{2}+\frac{c}{c+a}-<!--\; -\; -->\frac{1}{2}\ge <!--\; \ge \; -->0$
$<=>\sum <!--\; \sum \; -->\frac{a-<!--\; -\; -->b}{2(a+b)}\ge <!--\; \ge \; -->0$
$<=>x{y}^2+y{z}^2+x{z}^2-<!--\; -\; -->x^{2}y-<!--\; -\; -->y^{2}z-<!--\; -\; -->z^{2}x\ge <!--\; \ge \; -->0$
$<=>(y-<!--\; -\; -->x)(z-<!--\; -\; -->y)(z-<!--\; -\; -->x)\ge <!--\; \ge \; -->0$
And we have $y-<!--\; -\; -->x\le <!--\; \le \; -->0;z-<!--\; -\; -->x\le <!--\; \le \; -->0;z-<!--\; -\; -->y\ge <!--\; \ge \; -->0$
we are done !?

What can you deduce from $\cos <!--\; \; -->X(A-<!--\; -\; -->B)=0$?
So this is the problem:
$\sin <!--\; \; -->2x=\sqrt{2}\cos <!--\; \; -->x$
$\cos <!--\; \; -->x(2\sin <!--\; \; -->x-<!--\; -\; -->\sqrt{2})=0$
My question is, how can I come with this to the conclusion that, according to my answer sheet, $\cos <!--\; \; -->x=0$ and $\sin <!--\; \; -->x=\frac{\sqrt{2}}{2}$ Is it just plugging in something until it works?

Group satisfaction rate
Students are partitioned into groups of equal size $m$ similar to Stable roommates problem except $m$ can be any natural number greater than 1. Each student ranks the other students in strict order of preferences.
For each person $i\in \{1,\dots ,n\}$ can rank all other students from the most preferred to the least preferred. These preferences can be thought of as $p_{ij}$, where $p_{ij}$ is rank (1 highest) of $j$ in $i$'s ordering.
I'm trying to calculate the weighted rate of satisfaction a person would get mathematically from being in a group that contains his preference ranked at n. The weighted rate could be calculated with a function $f(m,p_{ij})=?$
The satisfaction rate is a number between 0 and 1 that multiplied by 100 could be converted to percentages.
Each preference should be weighted differently. If the student's group contains one of his preferences ranked at for example $p_{ij}=3$ he wouldn't be satisfied as much as a preference ranked at $p_{ij}=1$
The ideal group for a person would consist of his preferences ranked at $p_{ij}=1$ to $p_{ij}=m-<!--\; -\; -->1$ (amount of students in a group except for $i$ himself) this would give him a satisfaction rate of 1 (meaning fully satisfied), or written mathematically:
$\underset{p_{ij}=1}{\overset{m-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}f(m,p_{ij})=1$
So what could the definition of function $f(m,p_{ij})$ be?

What is a number between 1 and 50 that has at least 8 factors? write the answer as a fraction

How do I solve this fraction question?
If $a=-<!--\; -\; -->1/5$, how do I calculate:
$3a+2a^2$
I did $3\times <!--\; \times \; -->(-<!--\; -\; -->1/5)+(-<!--\; -\; -->1/5)\times <!--\; \times \; -->(-<!--\; -\; -->1/5)\times <!--\; \times \; -->2$, but can't figure out what the right way to solve this is.

Determine whether the Sequence is decreasing or increasing.
I have the sequence $\frac{({10}^n)}{(2n)!}$ and am trying to determine whether the sequence decreases or increases. I feel like the best way to proceed would be to use the squeeze theorem, but am unsure how to apply it to the problem.

What is the formula for the decimal representation of $\frac{a}{b}$ where $b$ is not coprime to 10?
If we want to turn a proper fraction $\frac{a}{b}$ into a decimal, then when $b$ is coprime to 10 we just need to rewrite $\frac{a}{b}$ in the form
$\frac{m}{{10}^{\mathrm{\varphi <!--\; \varphi \; -->}(b)}-<!--\; -\; -->1}$
where $\mathrm{\varphi <!--\; \varphi \; -->}$ is Euler's totient function. And then the repeatend is $m$ and the period of repetition is $\mathrm{\varphi <!--\; \varphi \; -->}(b)$.
But my question is, what do we do when $b$ is not coprime to 10? In that case we would need to rewrite $\frac{a}{b}$ in the form
$\frac{k+\frac{l}{m}}{{10}^n}$
where $k<{10}^n$, $\frac{l}{m}$ is a proper fraction, and $m$ is coprime to 10. Then $k$ would be the non-repeating part, $n$ would be the length of the nonrepeating part, and we can convert $l/m$ into a decimal using the procedure given in the beginning of my post.
But is there a formula for the four numbers $k$, $l$, $m$, and $n$ in terms of $a$ and $b$, short of doing long division to convert the fraction into a decimal?

Prove that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\le <!--\; \le \; -->1$
If $abc=1$ then
$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\le <!--\; \le \; -->1$
I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

How do I represent the following fraction in decimal?
I would like to know how do I represent the following fraction represented in base-$8$ in decimal form:
$(\frac{35}{25}{)}_{\text{8}}$
It seems very hard. Please somebody help me.

For which natural numbers a fraction can be simplifed
I want to know how to check for which natural numbers fraction $\frac{7n+19}{7n+2}$ can be simplified. The correct answer is supposed to be $n=2k-<!--\; -\; -->1$
I do understand that the denominator and the numerator have to have a common divisor in order to simplify the fraction, but fail to see how this particular problem should be approached.

How is symmetry in an inequality determined?
I was reading a book about inequalities, in that I found that
$$\begin{array}{}\text{(1)}& \frac{a}{b}+\frac{b}{c}+\frac{c}{a}>1\end{array}$$
is a symmetric inequality in a,b,c. But if I change the order from (a,b,c) to (a,c,b), I am not getting the same inequality which is the basic definition of symmetric inequality. What am I doing wrong?

Proof that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge <!--\; \ge \; -->a+b+c$
I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful.

How should this fraction involving powers be solved
$\sqrt{5}\cdot <!--\; \cdot \; -->{\left(\frac{5}{4}\right)}^{\frac{1}{2}}$
All I know is that this can be written as
$\sqrt{5}\cdot <!--\; \cdot \; -->{\left(\frac{5}{2^2}\right)}^{\frac{1}{2}}$
Any Ideas?

Improper fraction
Do the improper fractions include, "the fraction whose numerator is greater than or equal its denominator" or "the fraction whose numerator is greater only than the denominator"?

Rationalsing and simplifying surds question grade 10
I am a student and I am having difficulty answering this rationalising and simplifying question: I don't know what to do with the 2 at the bottom as a fraction.
$\frac{a}{\frac{\sqrt{40}}{2}}$
Thank you and help is appreciated