Explore Fraction Concepts with Examples and Questions

How to sum up these fractions?
Found this question in a competitive math test for elementary students. The long way is to add all the decimal values but is there a pattern/trick to solve this question (or these types)? I don't know how to solve this except by the long method of adding all the decimal equivalents.
The Answer is 1
Compute:
$\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{14}+\frac{1}{15}+\frac{1}{18}+\frac{1}{22}+\frac{1}{24}+\frac{1}{28}+\frac{1}{33}=?$

Negative Indices
I am not exactly good at evaluating negative indices--can someone please show me how to work out this expression:
$\frac{m^{-<!--\; -\; -->3}{n}^{-<!--\; -\; -->2}}{m^{-<!--\; -\; -->5}{n}^6}$
Both m's and the top n have negative indices.
Thanks!

How to simplify a diabolical expression involving radicals
A friend and I have been working on this problem for hours - how can the following expression be simplified analytically?
It equals $\frac{1}{2},$, and we have tried the following to no avail:
1. Substitution of $x=\sqrt{5}$
2. Substitution of $x=2\sqrt{5}$
3. Substitution of $x=5+\sqrt{5}$
4. Substitution of $x=\sqrt{5+\sqrt{5}}$
5. Manipulations by substituting the golden ratio
Here goes:
${\displaystyle \frac{{\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{2}}+{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}-<!--\; -\; -->{\displaystyle \frac{\sqrt{10+2\sqrt{5}}}{8}}}{{\displaystyle \frac{\sqrt{5(5+2\sqrt{5})}}{4}}+5\cdot <!--\; \cdot \; -->{\displaystyle \frac{\sqrt{5+2\sqrt{5}}}{4}}}}$
Thanks in advance for any help.

Equivalent forms of expressions with complex numbers
Which expressions are equivalent to $\frac{1}{(9i+z{)}^4}+\frac{1}{(9i-<!--\; -\; -->z{)}^4}$
Select all that apply.
$\frac{18i}{(81-z{)}^8}$
$\frac{-18i}{(81+z{)}^8}$
$\frac{18i}{(81+z{)}^8}$
$\frac{-18i}{(81-z{)}^8}$
Could someone walk me through this thoroughly? I would be incredibly grateful.

If $\frac{1}{x}-<!--\; -\; -->\frac{1}{y}=\frac{1}{z}$, $d=gcd(x,y,z)$ then $dxyz$ and $d(y-<!--\; -\; -->x)$ are squares
Let $x,y,z$ be three non negative integer such that ${\displaystyle \frac{1}{x}}-<!--\; -\; -->{\displaystyle \frac{1}{y}}={\displaystyle \frac{1}{z}}$. Denote by $d$ the greatest common divisor of $x,y,z$
Prove that $dxyz$ and $d(y-<!--\; -\; -->x)$ are squares number.
My idea is that if $x=da,y=db,z=dc$, where $gcd(a,b,c)=1$, then $c(b-<!--\; -\; -->a)=ba$. But I don't know how to do next.

limit of floor function, ceiling function and fractions
We know that
$\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->$ : greatest integer $\le <!--\; \le \; -->x$
$\lceil <!--\; \lceil \; -->x\rceil <!--\; \rceil \; -->$ : least integer $\ge <!--\; \ge \; -->x$
$\mathrm{f}\mathrm{r}\mathrm{a}\mathrm{c}(x)=x-<!--\; -\; -->\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->$
I can solve the questions limit of function like
$$\begin{gathered} \underset{x\to <!--\; \to \; -->n^{\pm <!--\; \pm \; -->}}{lim}\frac{\lfloor <!--\; \lfloor \; -->x-<!--\; -\; -->1\rfloor <!--\; \rfloor \; -->}{x-<!--\; -\; -->1} \\ \underset{x\to <!--\; \to \; -->n^{\pm <!--\; \pm \; -->}}{lim}\frac{\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->}{x-<!--\; -\; -->1} \end{gathered}$$
As $x$ approaches $n$ from above, $\lfloor <!--\; \lfloor \; -->x-<!--\; -\; -->1\rfloor <!--\; \rfloor \; -->=n-<!--\; -\; -->1$; therefore,
$\underset{x\to <!--\; \to \; -->n^{+}}{lim}\frac{\lfloor <!--\; \lfloor \; -->x-<!--\; -\; -->1\rfloor <!--\; \rfloor \; -->}{x-<!--\; -\; -->1}=\frac{n-<!--\; -\; -->1}{n-<!--\; -\; -->1}$
but I can't solve the questions like
$$\begin{gathered} \underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\lfloor <!--\; \lfloor \; -->x-<!--\; -\; -->3\rfloor <!--\; \rfloor \; -->}{x-<!--\; -\; -->1}, \\ \underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\lceil <!--\; \lceil \; -->x-<!--\; -\; -->3\rceil <!--\; \rceil \; -->}{x-<!--\; -\; -->1}, \\ \underset{x\to <!--\; \to \; -->a}{lim}\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->, \\ \underset{x\to <!--\; \to \; -->a}{lim}\lceil <!--\; \lceil \; -->x\rceil <!--\; \rceil \; -->, \\ \underset{x\to <!--\; \to \; -->a}{lim}\mathrm{f}\mathrm{r}\mathrm{a}\mathrm{c}(x) \end{gathered}$$
for $a\in <!--\; \in \; -->\mathbf{R}$
All questions are similar type, so I have given many problems in my questions. Please help me.

Is there a name for the rule $a÷<!--\; ÷\; -->(b\times <!--\; \times \; -->c)=a÷<!--\; ÷\; -->b÷<!--\; ÷\; -->c$?
Edit, because I should have looked it up before I posted the question:
Is there a name for the rule $a÷<!--\; ÷\; -->(b÷<!--\; ÷\; -->c)=a÷<!--\; ÷\; -->b\times <!--\; \times \; -->c$? I ran across this in Liping Ma's book, Knowing and Teaching Mathematics, and I have searched the internet for a name for this rule to no avail. It is not the distributive law, but it is rather similar. Thank you!
From Ma's book, p. 59 discussing "dividing by a number is equivalent to multiplying by its reciprocal":
"We can use the knowledge that students have learned to prove the rule that dividing by a fraction is equivalent to multiplying by its reciprocal. They have learned the commutative law. They have learned how to take off and add parentheses. They have also learned that a fraction is equivalent to to the result of a division, for example, $\frac{1}{2}=1÷<!--\; ÷\; -->2$ . Now, using these, we can rewrite the equation this way:
$1\frac{3}{4}÷<!--\; ÷\; -->\frac{1}{2}\to <!--\; \to \; -->$
$1\frac{3}{4}÷<!--\; ÷\; -->(1÷<!--\; ÷\; -->2)\to <!--\; \to \; -->$
$1\frac{3}{4}÷<!--\; ÷\; -->1\times <!--\; \times \; -->2\to <!--\; \to \; -->$ (This is the step my question is about.)
$1\frac{3}{4}\times <!--\; \times \; -->2÷<!--\; ÷\; -->1\to <!--\; \to \; -->$ (and I'd like an explicit explanation of this step, too.)
$1\frac{3}{4}\times <!--\; \times \; -->2\to <!--\; \to \; -->$$1\frac{3}{4}\times <!--\; \times \; -->(2÷<!--\; ÷\; -->1)\to <!--\; \to \; -->$

Finding out how out how much is 100%
I know 14% is 41. So how much is 100%? I know it is simple math, but... you know how it is being out of school for a few years. If at all possible, I'd like to see the equation too.

Why do I get $\mathrm{0.098765432098765432...}$ when I divide 8 by 81?
I got this remarkable thing when I divided 16 by 162, or, in a simplified version, 8 by 81. It's $0.098765432098765432\cdots <!--\; \cdots \; -->$ , or more commonly known as $0.\stackrel{¯<!--\; ¯\; -->}{098765432}$ with all the one-digit numbers going backwards...except for 1. Yeah, it's missing the 1. One, how do I get this remarkable outcome and two, why is it missing the 1?

How come ${\left(\frac{n+1}{n-<!--\; -\; -->1}\right)}^n={(1+\frac{2}{n-<!--\; -\; -->1})}^n$?
I'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:
${\left(\frac{n+1}{n-<!--\; -\; -->1}\right)}^n={(1+\frac{2}{n-<!--\; -\; -->1})}^n$
Except I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.
What am I missing?
Thanks.
Edit: Thanks to everyone who answered. Is there an "I feel stupid" badge? I really should have seen this a mile a way.

Elementary Fractions
There are two identical water jugs, A and B. Jug A is 3/7 full of water and Jug B is 8/11 full. What fraction of the capacity of a jug should water be poured out from jug B to jug A so that they both have the same amount of water?
Simple math, but not to me I tried many ways but to no avail (or i'm perhaps just dense).
The answer is 23/154
Please list down the steps and explain. Thank you!

Writing a fraction as $x^n$
I came across this fraction after practising with bunch for a while. How do I write this fraction
$\frac{1}{x^a}$
as
$x^n$
What happens to the $a$? I'm confused.

Why does the least common denominator work?
Take for instance the following problem. You have two beakers of the same height. One has tick marks that break it into thirds. The other has tick marks that separate it into fourths. The water levels are 1/3 and 1/4 respectively. If I did not know about the concept of LCDs, how would I figure out how much water there is all together? Please walk me through your reasoning.
Note: I understand the need to find a common scale between the two beakers. I don't know how I would find that 12 is the smallest possible common scale, if I had never been introduced to the concept of LCDs/LCMs.

Find the integer closest to $a-<!--\; -\; -->b$
Let
$a=\frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\dots <!--\; \dots \; -->+\frac{{1001}^2}{2001}.$
Let
$b=\frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\dots <!--\; \dots \; -->+\frac{{1001}^2}{2003}.$
Then, what is the closest integer of $(a-<!--\; -\; -->b)$?

$-<!--\; -\; -->\frac{7}{6}=-<!--\; -\; -->\frac{4n}{3}$ struggling on solving this equation
I was wondering how to do this equation step by step. I forgot how to but it would be awesome if someone could help me out
$-<!--\; -\; -->\frac{7}{6}=-<!--\; -\; -->\frac{4n}{3}$

Resolve $\frac{120}{x+y}+\frac{60}{x-<!--\; -\; -->y}=6;\frac{80}{x+y}+\frac{100}{x-<!--\; -\; -->y}=7$
I want to resolve this system of equations:
$\{\begin{array}{l}\frac{120}{x+y}+\frac{60}{x-<!--\; -\; -->y}=6\\ \frac{80}{x+y}+\frac{100}{x-<!--\; -\; -->y}=7\end{array}$I came to equations like
$x-<!--\; -\; -->\frac{10x}{x-<!--\; -\; -->y}+y-<!--\; -\; -->\frac{10y}{x-<!--\; -\; -->y}=20$
and
$-<!--\; -\; -->2xy-<!--\; -\; -->y^2-<!--\; -\; -->10y=20-<!--\; -\; -->x^2-<!--\; -\; -->10x$
I need to leave x or y alone and didn't succeed. Any help?

What is the fastest method to find which of $\frac{3\sqrt{3}-<!--\; -\; -->4}{7-<!--\; -\; -->2\sqrt{3}}$ and $\frac{3\sqrt{3}-<!--\; -\; -->8}{1-<!--\; -\; -->2\sqrt{3}}$ is bigger manually?

Compare colon notation with fraction
I'm working on a job interview test and there is one answer which I just don't get.
The test states that statement below is true. To me it just seems wrong. No box is provided to check. Then how do I check it correct or wrong? Am I missing something here?
66:4 = 161/2

Prove that $\frac{1}{1+a_1+a_1{a}_2}+\frac{1}{1+a_2+a_2{a}_3}+\cdots <!--\; \cdots \; -->+\frac{1}{1+a_{n-<!--\; -\; -->1}+a_{n-<!--\; -\; -->1}{a}_n}+\frac{1}{1+a_n+a_n{a}_1}>1.$
I find it hard to use any inequalities here since we have to prove $>1$ and most inequalities such as AM-GM and Cauchy-Schwarz use $\ge <!--\; \ge \; -->1$. On the other hand it seems that if I can prove that each fraction is $>1$ that might help, but I am unsure.

Inequality - GM, AM, HM and SM means
I've got stuck at this problem :
Prove that for any $a>0$ and any $b>0$ the following inequality is true:
$3(\frac{a^3}{b^3}+\frac{b^3}{a^3})\ge <!--\; \ge \; -->\frac{a}{b}+\frac{b}{a}+4$
The first thing that I've thought was the AM-GM inequality (the extended version - heard that is also known as The power mean inequality):
$HM\le <!--\; \le \; -->GM\le <!--\; \le \; -->AM\le <!--\; \le \; -->SM$
where $HM$, $GM$, $AM$, and $SM$ refer to the harmonic, geometric, arithmetic, and square mean, respectively. CBS(Cauchy - Buniakowsky - Schwartz) also come to my mind, but I think it isn't helpful in this case.
I would be greatful for some hints.
Thanks!