Explore Fraction Concepts with Examples and Questions

Is $(x^2+1)/(x^2-<!--\; -\; -->5x+6)$ divisible?
I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate
$\frac{x^2+1}{x^2-<!--\; -\; -->5x+6}.$
I know it is improper rational function and to make it proper rational fraction we have to divide
$\frac{x^2+1}{x^2-<!--\; -\; -->5x+6}$
I'm trying from sometime but couldn't find the right solution.
Please help! Thank you in advance.

A question about how to express a fraction as $\frac{1}{q_1}+\frac{1}{q_2}+\cdots <!--\; \cdots \; -->+\frac{1}{q_N}$
Let x be a positive rational number, strictly between 0 and 1. Prove that there is a finite strictly increasing list of positive integers $2\le <!--\; \le \; -->q_1<q_2<\cdots <!--\; \cdots \; --><q_N$ such that
$x=\frac{1}{q_1}+\frac{1}{q_2}+\cdots <!--\; \cdots \; -->+\frac{1}{q_N}.$
I have tried many methods, such as mathematical induction. I know it is obvious that $\frac{1}{m}$ is right for assumption, but when I begin to prove $\frac{2}{m}$ I feel it is hard to find some patterns for $\frac{2}{m}$. Thus, maybe my thoughts are wrong.
Also, I still tried to change the fraction so that the numerator will be smaller and smaller. But I still cannot find a way to lower the numerator. Can someone help me solve the question? Or, can someone give me some hints. I will appreciate you very much!

Removing additive factors from the denominator of a fraction
Suppose I have a variable $x_L$ defined as follows:
$x_L=\frac{r_1{e}^{-<!--\; -\; -->t_{0}s}}{s+r_2+r_3},$
Is there a way for me to rearrange this fraction to get something in this form
$x_L=[\mathrm{stuff}]\dots <!--\; \dots \; -->\frac{[\mathrm{stuff}]}{s},$
where "$\dots <!--\; \dots \; -->$" is any operator (e.g., $+$ or $\times <!--\; \times \; -->$)?

Distributing multiplication of rational functions
I am having trouble distributing with fractions. This
$(\frac{1}{(x+3)}+\frac{(x+3)}{(x-<!--\; -\; -->3)})(9-<!--\; -\; -->x^2)=-<!--\; -\; -->\frac{(x^2-<!--\; -\; -->3)}{9-<!--\; -\; -->x^2}(9-<!--\; -\; -->x^2)$
has the answer $\left\{[x=-<!--\; -\; -->\frac{9}{7}]\right\}$
I start solving by cancelling $(9-<!--\; -\; -->x^2)$ on the right hand side, and multiplying $(9-<!--\; -\; -->x^2)$ with the terms on the left hand side by foiling (multiplying term by term). So the first step is
$9\times <!--\; \times \; -->\frac{1}{x+3}+\frac{9(x+3)}{x-<!--\; -\; -->3}-<!--\; -\; -->x^2\times <!--\; \times \; -->\frac{1}{x+3}-<!--\; -\; -->\frac{x^2(x+3)}{x-<!--\; -\; -->3}=-<!--\; -\; -->x^2-<!--\; -\; -->3$
This is apparently wrong because it has another answer, $\left\{[x=-<!--\; -\; -->\frac{3}{7}]\right\}$ What am I doing wrong?

Inequality with the constraint $xy=x+y$
I have to prove that if $x$ and $y$ are two positive numbers such that $xy=x+y$, the following inequality
$\frac{x}{y^2+4}+\frac{y}{x^2+4}\ge <!--\; \ge \; -->\frac{1}{2}$
holds. Any help is appreciated, thank you in advance.

How to show $\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{3}{4}\cdot <!--\; \cdot \; -->\frac{5}{6}\cdots <!--\; \cdots \; -->\frac{99}{100}<\frac{1}{12}$?

Inequality for an alternating sum of binomial coefficients times a fraction
I need to show that
$\underset{n=0}{\overset{x}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^n(\genfrac{}{}{0ex}{}{x}{n})(\frac{1}{n+l-<!--\; -\; -->1}-<!--\; -\; -->\frac{1}{n+l})>0,$
for any $l>0$. I tried to prove this, but I didn't get anywhere.
Apparently the above sum is equal to $\frac{(l-<!--\; -\; -->2)!(x+1)!}{(x+l)!}$. However I have no idea how to prove this either. Any tips on how to approach this would be greatly appreciated!

Express the following product as a single fraction:
$(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots <!--\; \cdots \; -->(1+\frac{1}{3^{(2n)}})$
I'm having difficulty with this problem:
What i did was:
I rewrote the $1$ as $\frac{3}{3}$
here is what i rewrote the whole product as:
$(\frac{3}{3}+\frac{1}{3})(\frac{3}{3}+\frac{1}{3^2})(\frac{3}{3}+\frac{1}{3^4})\cdots <!--\; \cdots \; -->(\frac{3}{3}+\frac{1}{3^{2^n}})$
but how would i proceed after this?

How to solve complex fractions?
$f\circ <!--\; \circ \; -->g=\frac{{\displaystyle \frac{x+3}{x-<!--\; -\; -->6}}-<!--\; -\; -->2}{{\displaystyle \frac{x+3}{x-<!--\; -\; -->6}}+8}$
How would I solve this complex fraction? I know what the answer is, but I am just not sure how they got there. The answer is
$\frac{-<!--\; -\; -->x+15}{9x-<!--\; -\; -->45}$
I have tried multiplying both sides by $(x-<!--\; -\; -->6)$ but I am getting $x^2-<!--\; -\; -->3x-<!--\; -\; -->18$? What am I doing wrong here?

Simplify the fraction with radicals
I want to simplify this fraction
$\frac{\sqrt{6}+\sqrt{10}+\sqrt{15}+2}{\sqrt{6}-<!--\; -\; -->\sqrt{10}+\sqrt{15}-<!--\; -\; -->2}$
I've tried to group up the denominator members like $(\sqrt{6}+\sqrt{15})-<!--\; -\; -->(\sqrt{10}+2)$ and then amplify with $(\sqrt{6}+\sqrt{15})+(\sqrt{10}+2)$

Find the value of x
$(1-<!--\; -\; -->\frac{1}{2^2})(1-<!--\; -\; -->\frac{1}{3^2})(1-<!--\; -\; -->\frac{1}{4^2})\dots <!--\; \dots \; -->(1-<!--\; -\; -->\frac{1}{{2011}^2})=\frac{x}{2\cdot <!--\; \cdot \; -->2012}$
Help me solve this, obviously there is a shortcut to solve it, but I have tried a lot and I've come up with something confusing!

Is $\frac{\sqrt{7}}{\sqrt[3]{15}}$ rational or irrational? Prove it.
I am having a hard time with this question. So far what I did was say, assume it's rational, then
$\frac{\sqrt{7}}{\sqrt[3]{15}}=\frac{x}{y}\Rightarrow <!--\; \Rightarrow \; -->\sqrt{7}y=x\sqrt[3]{15}$
I then showed the product of a rational number and an irrational number is irrational so the expression above is irrational on the left and right side. I can't get to a contradiction to prove it's irrational so I'm currently thinking it is rational but I don't know how to go about proving it.

How is it possible that when you divide 1 by 9,899, you get two-digit Fibonacci numbers also being carried, etc.?
When I divided 1 by 9,899, I got two-digit Fibonacci numbers also being carried: $0.0001010203050813213455904636\dots <!--\; \dots \; -->$
When I divided 1 by 89, I got one-digit Fibonacci numbers at the beginning: 0.0112359… (there was originally an eight, but there was carrying and it changed to a nine)
How does all of this happen? There is more than this, you know. What do you think is going on here?

ln() and product confuse me
I have this function:
$p(x_n;\mathrm{\mu <!--\; \mu \; -->})=\frac{1}{(2\mathrm{\pi <!--\; \pi \; -->}{)}^{\frac{l}{2}}|\mathrm{\Sigma <!--\; \Sigma \; -->}{|}^{\frac{1}{2}}}\exp <!--\; \; -->(-<!--\; -\; -->\frac{1}{2}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^T{\mathrm{\Sigma <!--\; \Sigma \; -->}}^{-<!--\; -\; -->1}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}))$
And I have to compute this: $L(\mathrm{\mu <!--\; \mu \; -->})=\ln <!--\; \; -->(\underset{n=1}{\overset{N}{\prod <!--\; \prod \; -->}}p(x_n;\mathrm{\mu <!--\; \mu \; -->}))$ which should be equal to:
$-<!--\; -\; -->\frac{N}{2}\ln <!--\; \; -->((2\mathrm{\pi <!--\; \pi \; -->}{)}^l|\mathrm{\Sigma <!--\; \Sigma \; -->}|)-<!--\; -\; -->\frac{1}{2}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^T{\mathrm{\Sigma <!--\; \Sigma \; -->}}^{-<!--\; -\; -->1}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})$
Note: $\mathrm{\Sigma <!--\; \Sigma \; -->}$ is a covariance matrix.
From my attempt I have:
$\frac{1}{(2\mathrm{\pi <!--\; \pi \; -->}{)}^{\frac{l}{2}}|\mathrm{\Sigma <!--\; \Sigma \; -->}{|}^{\frac{1}{2}}}=\ln <!--\; \; -->1-<!--\; -\; -->\ln <!--\; \; -->((2\mathrm{\pi <!--\; \pi \; -->}{)}^{l/2}|\mathrm{\Sigma <!--\; \Sigma \; -->}{|}^{1/2})=-<!--\; -\; -->\ln <!--\; \; -->((2\mathrm{\pi <!--\; \pi \; -->}{)}^{l/2}|\mathrm{\Sigma <!--\; \Sigma \; -->}{|}^{1/2})=-<!--\; -\; -->\frac{L}{2}\ln <!--\; \; -->2\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\frac{1}{2}\ln <!--\; \; -->|\mathrm{\Sigma <!--\; \Sigma \; -->}|$
and:
$\ln <!--\; \; -->(\exp <!--\; \; -->(-<!--\; -\; -->\frac{1}{2}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^T{\mathrm{\Sigma <!--\; \Sigma \; -->}}^{-<!--\; -\; -->1}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})))=-<!--\; -\; -->\frac{1}{2}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^T{\mathrm{\Sigma <!--\; \Sigma \; -->}}^{-<!--\; -\; -->1}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})$
but when I put everything together with the product, it doesn't work out, squares appear and it doesn't feel right...Can you help me please understand?

Basic solving for fractions
Can someone help me with this? I am a beginner to physics and in a question i need no isolate a constant to solve for another one.
A/B=C/D solve for D step by step please.

How do I express 14/(3-√2) in the form of b+c√d?
I don't really understand the method to express a surd. Can anyone help me with this question? I really want more ideas on how to solve it, maybe there is one I may understand well. TY

Shorten $\frac{n}{n^{\frac{1}{2}}}$?I have a short question to solve my problem.
Can I simplify $\frac{n}{n^{\frac{1}{2}}}$?
Thanks already for answers.

How can I integrate $\int <!--\; \int \; -->\frac{1}{2x+2}$
Method 1
$\int <!--\; \int \; -->\frac{1}{2x+2}=\frac{1}{2}\int <!--\; \int \; -->\frac{1}{x+1}=\frac{1}{2}ln(x+1)+c$
Method 2
$\int <!--\; \int \; -->\frac{1}{2x+2}=\frac{1}{2}\int <!--\; \int \; -->\frac{2}{2x+2}=\frac{1}{2}ln(2x+2)+c$
Wolfram Alpha suggests the first method yields the correct answer, but why do the two methods produce different answers?

Adding fractions of Groups of People
I understand the rules of adding fractions perfectly well. I know how to find common denominators, and understand why adding fractions without common denominators doesn't make sense.
But, today someone asked me about adding $\frac{5}{6}$ and $\frac{21}{28}$. They were wondering why dividing the two factions and taking the average ($\frac{0.8333+0.75}{2}=0.7917$) was giving them a different value than adding them together, and then dividing ($\frac{5}{6}+\frac{21}{28}=\frac{26}{34}=0.7647$).
My initial reaction was the same as yours: I was taken aback by someone adding fractions in this way, and I gave a quick refresher of why this doesn't make sense, and how to properly add fractions.
I thought I had helped them fix their problem, until they gave more more context: The $\frac{5}{6}$ was five people in a group of six who were observed washing their hands after a certain activity. The $\frac{21}{28}$ was twenty-one out of twenty-eight people who were observed washing their hands after a certain activity. The goal was to find the total number of people who had washed their hands, as a fraction of the total number of observed people. So, $\frac{26}{34}$ is actually correct in this case.
But, I'm still having trouble reconciling this with what I know about fractions. At least, what I think I know. Is there a term for these sorts of fractions? This isn't like having five slices of a six-slice pizza combined with twenty-one slices of a twenty-eight-slice pizza. This is like having one oven holding six pizzas, five of which have mushrooms, and another (huge) oven holding twenty-eight pizzas, twenty-one of which have mushrooms. What fraction of the pizzas have mushrooms? Is it $\frac{5}{6}+\frac{21}{28}$ ? I don't think so.
Is there some terminology I'm forgetting here? Why am I getting so tripped up by this?

Simplify rational fraction
$\frac{bx(a^2{x}^2+2a^2{y}^2+b^2{y}^2)+ay(a^2{x}^2+2b^2{x}^2+b^2{y}^2)}{bx+ay}$
My attempt:
${\displaystyle \frac{b{a}^2{x}^3+2a^{2}bx{y}^2+b^{3}x{y}^2+a^3{x}^{2}y+2b^2{x}^{2}ay+b^{2}a{y}^3}{bx+ay}}$
I multiplied it out like this, then got stuck...