Explore Fraction Concepts with Examples and Questions

Find the fractional representation $p/q$
Been trying to get some sort of solution for this for hours now, with no avail.
Find the fractional representation $p/q$, with $p\in <!--\; \in \; -->\mathbb{N}$ and $q\in <!--\; \in \; -->\mathbb{N}$, of the rational number whose decimal representation is:
$\mathrm{22.521111...}$
Any help would be greatly appreciated.
Thanks

Monotonicity of a fraction.
So I want to prove that the following fraction is monotone increasing, as a part of another proof, that's why I stumbled on:
$\frac{4^{n+1}}{2\sqrt{n+1}}\ge <!--\; \ge \; -->\frac{4^n}{2\sqrt{n}}$
I know it's basic, though how to prove it?

Comparing two fractions
I saw this problem from an elementary textbook:
Let
$A=\frac{2014}{2015}+\frac{2015}{2016}$
and
$B=\frac{2014+2015}{2015+2016}$
Compare A and B.
I know the answer is $A><!--\; >\; -->B$ because
$\frac{A}{B}><!--\; >\; -->1.$
But I cannot find a precise way to explain it. Can anyone please help? Thank you.

How to solve ${\displaystyle \frac{7x}{8}}+4-<!--\; -\; -->{\displaystyle \frac{2x}{3}}=4x-<!--\; -\; -->3$
I do not understand how to simplify this. Could anyone here help me, please? Thanks.

Hidden Cauchy-Schwarz inequality
If $x_i><!--\; >\; -->0$ and $x_i{y}_i-<!--\; -\; -->{z_i}^2><!--\; >\; -->0$ for $i\le <!--\; \le \; -->n$ , then prove that
$\frac{n^3}{(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i)(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i)-<!--\; -\; -->(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{z_i}^2)}\le <!--\; \le \; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{x_i{y}_i-<!--\; -\; -->z_i^2}$
This is a question from the IMO 1969. At first it seems that we can assume that $a_i=\sqrt{x_i{y}_i}-<!--\; -\; -->z_i$ and $b_i=\sqrt{x_i{y}_i}+z_i$. Hence RHS becomes $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{a_i{b}_i}$.Now how do we simplify the LHS so that we can use the Cauchy-Schwarz inequality?

Prove or disprove that $|a_1|+|a_2|+\dots <!--\; \dots \; -->+|a_n|\le <!--\; \le \; -->n\sqrt{a_1^2+\dots <!--\; \dots \; -->+a_n^2}$ by showing that $RHS-<!--\; -\; -->LHS\ge <!--\; \ge \; -->0$if possible.

What number is midway between 2/4 and 3/4?
Sorry for posting this question but appreciate if you could help me.
What number is midway between 2/4 and 3/4?

Are the sums equal to each other?
They are $2$ different results for the integral
$\int <!--\; \int \; -->x{e}^{2x}\sin <!--\; \; -->\left(\frac{x}{3}\right)dx$
1. ${\displaystyle \frac{-<!--\; -\; -->3}{1369}{e}^{2x}(3(35-<!--\; -\; -->74x)\sin <!--\; \; -->\left(\frac{x}{3}\right)+(37x-<!--\; -\; -->36)\cos <!--\; \; -->\left(\frac{x}{3}\right))}$
2. ${\displaystyle \frac{3}{37}{e}^{2x}(x(6\sin <!--\; \; -->\left(\frac{x}{3}\right)-<!--\; -\; -->\cos <!--\; \; -->\left(\frac{x}{3}\right))-<!--\; -\; -->\frac{3}{37}(35\sin <!--\; \; -->\left(\frac{x}{3}\right)-<!--\; -\; -->13\cos <!--\; \; -->\left(\frac{x}{3}\right)))}$
Are the sums equal to each other?

Summation and Patterns Question
I have a really urgent question. Today i was looking at this Pattern (Using fractions) :
$\frac{1}{1\ast <!--\; \ast \; -->3}+\frac{1}{3\ast <!--\; \ast \; -->5}+\frac{1}{5\ast <!--\; \ast \; -->7}+\frac{1}{7\ast <!--\; \ast \; -->9}+...$=? The pattern is $\frac{n}{2n+1}$. I was wondering if you could do a summation formula, in which the end term is infinite. Thanks for the help!

Rationalise the denominator and simplify $\frac{3\sqrt{2}-<!--\; -\; -->4}{3\sqrt{2}+4}$
Does someone have an idea how to work ${\displaystyle \frac{3\sqrt{2}-<!--\; -\; -->4}{3\sqrt{2}+4}}$ by rationalising the denominator method and simplifying?

Prove that if $a+b+c=1$ then $\underset{cyc}{\sum <!--\; \sum \; -->}\frac{1}{\sqrt{a^2+b^2}}\le <!--\; \le \; -->\frac{9\sqrt{2}}{2}$
My try:since
$\sqrt{a^2+b^2}\ge <!--\; \ge \; -->{\displaystyle \frac{\sqrt{2}}{2}}(a+b)$
it suffices to prove that
$\underset{cyc}{\sum <!--\; \sum \; -->}{\displaystyle \frac{1}{a+b}}\le <!--\; \le \; -->{\displaystyle \frac{9}{2}}$
But since Cauchy-Schwarz inequality we have
$2\underset{cyc}{\sum <!--\; \sum \; -->}(a+b)\underset{cyc}{\sum <!--\; \sum \; -->}{\displaystyle \frac{1}{a+b}}\ge <!--\; \ge \; -->9$
or
$\underset{cyc}{\sum <!--\; \sum \; -->}{\displaystyle \frac{1}{a+b}}\ge <!--\; \ge \; -->{\displaystyle \frac{9}{2}}$
so this try can't works.so How to prove $(1)$

Where did my simplification go wrong? Sum and difference formula simplification
I'm struggling with the following: We are to use the sum and difference formulas to find the exact value of the expression. The problem is simplification has been tough. As a last resort I decided to use Symbolab to find the answer and steps but the steps were not to be found. Despite lack of steps, the answer is $(\sqrt{2+\sqrt{3}})/2$
Here are my steps so far:
$\begin{array}{rl}\sin <!--\; \; -->({135}^{\circ <!--\; \circ \; -->}-<!--\; -\; -->{30}^{\circ <!--\; \circ \; -->})& =(\sin <!--\; \; -->{135}^{\circ <!--\; \circ \; -->}\cos <!--\; \; -->{30}^{\circ <!--\; \circ \; -->})-<!--\; -\; -->(\cos <!--\; \; -->{135}^{\circ <!--\; \circ \; -->}\sin <!--\; \; -->{30}^{\circ <!--\; \circ \; -->})\\ & =\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-<!--\; -\; -->(-<!--\; -\; -->\frac{\sqrt{2}}{2})\left(\frac{1}{2}\right)\\ & =\left(\frac{\sqrt{6}}{4}\right)-<!--\; -\; -->(-<!--\; -\; -->\frac{\sqrt{2}}{4})\end{array}$
[insert final simplification step]
$(\sqrt{2+\sqrt{3}})/2$
What's the missing step here?

If $a+b$,= 5 and $ab=6$ . Find $\frac{1}{a}$ + $\frac{1}{b}$
Any Ideas on how to begin?
Many Thanks

Help with some algebra
Can anybody explain to me how they went from this
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
to this.
$(y_1-y_2)·x-(x_1-x_2)·y-(y_1-y_2)·x_1+(x_1-x_2)·y_1=0$
Its confusing..

How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$
How can I simplify as much as possible:
$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$
Also, first post here, looking forward to sticking around!

Why does $\sqrt{x}/y=\sqrt{x/y/y}$
Sorry for the awkward title, hard to to sum a mathematical problem with words alone.
Having said that, I recently learned that the root of any value, x, and then that over value y, is identical to the root of $x/y/y$ , as in $\sqrt{x/y/y}$
For e.g, $\sqrt{35}/7=\sqrt{5/7}$ and I cannot logically deduce why. So far, I know that:
$$\begin{gathered} 35=7\times <!--\; \times \; -->5 \\ \frac{\sqrt{35}}{7}=\frac{\sqrt{7}\times <!--\; \times \; -->\sqrt{5}}{7} \end{gathered}$$
And somehow this being equal to:
√(5/7)
So it seems that we divided the numerator, √35 by √5 to give √7. However, 7/√7 is not √7! Yet, apparently it is?
So, a fraction is of course equal to itself if scaled up or down by the same amount, in that 10/2 is the same as 5/1, or 5. So scaling down √35 by √5 makes sense, as it only leaves √7, but it seems that the same cannot be said for the denominator, 7. 7/√7 = √7!? Yet, the two expressions really are the same, √35/7 and √(5/7).
Thanks for any help in advance.
P.S This works for any values, so it's no coincidence.

Convert fraction into decimal
Trying to help my daughter with a question for her maths that has got me stuck...wish I was better at maths!
Convert the fraction 27/50 into decimal
Could someone please help me with this question.

Find the limit of fraction involving logarithms
I am looking for a way to prove the following limit for integer $x$s:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\log <!--\; \; -->(x+2)-<!--\; -\; -->\log <!--\; \; -->(x+1)}{\log <!--\; \; -->(x+2)-<!--\; -\; -->\log <!--\; \; -->(x)}=\frac{1}{2}$
I could find the result by using a computer program but I cannot formally establish the above equality.

5.30 converted to a fraction or mixed number in lowest terms
The correct answer they got on my worksheet is $5\frac{3}{50}$, but I get $5\frac{15}{50}$

Prove inequality by induction $\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N},n\ge <!--\; \ge \; -->2:(1+\frac{1}{n}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}<3$
I have the following statement to prove:
$\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N},n\ge <!--\; \ge \; -->2:(1+\frac{1}{n}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}<3$
What I have already proven is that
$(1+\frac{1}{n}{)}^n<(1+\frac{1}{n+1}{)}^{n+1}$
I have started, as usual by induction with $n=2$. Then I went on to say
$(1+\frac{1}{n+1})\ast <!--\; \ast \; -->(1+\frac{1}{n+1}{)}^n<\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k!}+\frac{1}{(n+1)!}<3$
But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now $n+1$ instead of $n$. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?