Explore Fraction Concepts with Examples and Questions

I can't understand the algebraic simplification below
I can't understand the last step of simplification in the following algebraic expression.
I am aware that $\sqrt{3}=3^{\frac{1}{2}}$. Can't see how they get the $6$ in the denominator, and how $\sqrt{3}$ gets to the numerator. Any help will be appreciated.
p.s. this is from a mathematics text book. last step of a derivative calculation.
$=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}$

How to simplify $\sqrt{3/2}$
My textbook says to do the following steps to simplify the $\sqrt{3/2}$
$\sqrt{3/2}=\sqrt{((3/2)\times <!--\; \times \; -->(2/2))}=\sqrt{6/4}=\sqrt{6}/\sqrt{4}=\sqrt{6}/2.$
However, if the fourth step is valid (1. please explain why and 2. I specifically mean where you "split" the square root into two parts), then why can't you simply do:
$\sqrt{3/2}=\sqrt{3}/\sqrt{2}$
$\sqrt{3/2}=\sqrt{3}/\sqrt{2}$
Why wouldn't that be valid?
My point is, are the two fractions equivalent?

Comparison of two values
I have to figure out the relation between the quantity $(0.9/1.1{)}^2+(1.1/0.9{)}^2$ and 2. How can i do this without explicitly calculating the first value, by using some laws of exponents?

How to simplify this expression with radicals? $3\sqrt{2}-<!--\; -\; -->\sqrt{32}+\sqrt{\frac{80}{16}}$
My answer is $-<!--\; -\; -->\sqrt{2}+\sqrt{5}$ , but the real answer should be ${\displaystyle \frac{9-<!--\; -\; -->4\sqrt{2}}{4}}$

Question on recurring decimal digits
In my discrete maths class, I have come across an interesting phenomenon for which I can't find an explanation!
If we divide $1$ by $13$ we obtain $0.07692307\dots <!--\; \dots \; -->$
If we divide $3$ by $13$ we obtain $0.23076923\dots <!--\; \dots \; -->$
If we divide $4$ by $13$ we obtain $0.30769230\dots <!--\; \dots \; -->$
As you can see, the digits are recurring in the same order but starting at a different point in the sequence.
Can someone explain this to me? What exactly is happening here?

Find a common denominator
I'm trying to integrate a function but first I need to find a common denominator for:
$\frac{A}{x-<!--\; -\; -->8}$+$\frac{B}{x+1}$+$\frac{C}{x-<!--\; -\; -->1}$

Expressing as a single fraction
I'm trying to improve my maths so I can sit the GRE this time next year.
I'm stuck on a really silly question:
Express as a single fraction:
$\frac{\frac{3x}{2y-<!--\; -\; -->7y}}{4x}$
I'm trying to find a common denominator to cancel out the bottom line. Am I on the right track?

simple multiplication question: multiplying radial fractions
I'm having a bit of brain block right now.
In order to multiply any fractions you simply reduce to lowest form, then multiply num., then denom. When I tried with fractions in radian form it didn't work out so well.
consider: $(6\mathrm{\pi <!--\; \pi \; -->}/5)(2\mathrm{\pi <!--\; \pi \; -->}/3$
If I want to keep the $\mathrm{\pi <!--\; \pi \; -->}$ in the numerator I thought I should just multiply $6$ and $2$ to get $12$ for $12\mathrm{\pi <!--\; \pi \; -->}$. But I don't think that's a mathematically sound method of multiplication. So I figured I should just multiply the entire numerator and then divide out $\mathrm{\pi <!--\; \pi \; -->}$: so that's $(6\mathrm{\pi <!--\; \pi \; -->})(2\mathrm{\pi <!--\; \pi \; -->})=118.4352528$ then divide out $\mathrm{\pi <!--\; \pi \; -->}$ which gives me $37.69911184$ which isn't right because it should be an integer.I know I'm missing something very fundamental here. Any help appreciated.

Finding LCM of an expression
$\frac{x}{x^2-<!--\; -\; -->8x+7}-<!--\; -\; -->\frac{x}{x^2-<!--\; -\; -->2x-<!--\; -\; -->35}=$$\frac{6x}{(x-<!--\; -\; -->7)(x-<!--\; -\; -->1)(x+5)}$
I am currently trying to figure out how to solve this problem. I know how they got the denominator but am unsure how they got 6x in the numerator? Can somebody explain this to me?

To split in to partial fractions, the expression ${\displaystyle \frac{1}{x^2{(x+a)}^2}}$
${\displaystyle \frac{1}{x^2{(x+a)}^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{{(x+a)}^2}}$
One method of finding the values of the constants ${\displaystyle A,B,C\text{ \& }D}$ is as follows.
${\displaystyle Ax{(x+a)}^2+B{(x+a)}^2+C\left(x^2\right)(x+a)+D{x}^2=1}$
${\displaystyle Ax(x^2+a^2+2ax)+B(x^2+2ax+a^2)+C(x^3+a{x}^2)+D{x}^2=1}$
${\displaystyle A(x^3+a^{2}x+2a{x}^2)+B(x^2+2ax+a^2)+C(x^3+a{x}^2)+D{x}^2=1}$
${\displaystyle x^3(A+C)+x^2(2Aa+B+Ca+D)+x(A{a}^2+2Ba)+B{a}^2=1}$
Equating coeffecients of ${\displaystyle x^3,x^2,x^1,x^0}$
${\displaystyle A+C=0,2Aa+B+Ca+D=0,A{a}^2+2Ba=0,B{a}^2=1}$
${\displaystyle B=\frac{1}{a^2}}$
${\displaystyle A=\frac{-2}{a^3}}$
${\displaystyle C=\frac{2}{a^3}}$
${\displaystyle D=\frac{1}{a^2}}$
Is there a shorter method to find the values of A,B,C & D ?

Unknown terms of the proportion
please help me solving this problem. The question is, find the unknown terms of the proportion
${\displaystyle \frac{2}{3}=\frac{x}{12}=\frac{y}{15}.}$

How to simplify ${\displaystyle \frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}}$
I was tackling through an olympiad practice book when I saw one of these problems:
If ${\displaystyle x=5+2\sqrt{6}}$, evaluate ${\displaystyle \frac{x-1}{\sqrt{x}}}$
The answer written is ${\displaystyle 2\sqrt{2}}$, but I can't figure my way out through the manipulations. I just know that I have the following:
${\displaystyle \frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}}$

Compute 4*15*14-9*7 using the correct order of operations.

Compute 7/8*14-12/14/8 using the correct order of operations.

Evaluate 9-11/9 +4-8-11.

Reduce 4-10*15+10/14+5

Reduce 9+6*3+5*8-7.

Evaluate fraction of sum
So i have to evaluate this sum:
${\displaystyle \frac{1-2^{-2}+4^{-2}-5^{-2}+7^{-2}-8^{-2}+{10}^{-2}-{11}^{-2}+\cdots }{1+2^{-2}-4^{-2}-5^{-2}+7^{-2}+8^{-2}-{10}^{-2}-{11}^{-2}+\cdots }}$
it has the form : ${\displaystyle \frac{\sum _0^{\mathrm{\infty }}[{(3n+1)}^{-2}-{(3n+2)}^{-2}]}{\sum _0^{\mathrm{\infty }}{(-1)}^n[{(3n+1)}^{-2}+{(3n+2)}^{-2}]}}$
My current attempt : Trying to convert this into power series
${\displaystyle a\left(n\right)={(3n+1)}^{-2}-{(3n+2)}^{-2}b\left(n\right)={(3n+1)}^{-2}+{(3n+2)}^{-2}}$
Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ?
Maybe there is a better direction. Can someone give me a hint ?

Evaluate the expression
${\displaystyle \frac{1^2}{1^2-10+50}+\frac{2^2}{2^2-20+50}+\cdots +\frac{{80}^2}{{80}^2-80+50}}$
What should I do after that ?
${\displaystyle \sum \frac{n^2}{n^2-10n+50}}$
I'm not seeing anything to find some way to cancel out the terms or something like that !