Explore Fraction Concepts with Examples and Questions

Advice on simple exercise; dividing rational expression
I'm looking into rational expressions to clarify some doubts that i have. I'm solving the exercises at the end of the chapter and I'm stuck with one of them. This is it:
$$\frac{(x-<!--\; -\; -->y{)}^2-<!--\; -\; -->z^2}{(x+y{)}^2-<!--\; -\; -->z^2}÷<!--\; ÷\; -->\frac{x-<!--\; -\; -->y+z}{x+y-<!--\; -\; -->z}$$
I have been solving several of this type by:
1.- Factoring out the largest common factor if possible.
2.- Factoring.
3.- Removing factors of 1.
In this case, i'm trying to find a way of solving this without expanding the hole thing, but i have not found the way jet.
I start by expanding both square binomials of the dividend.
$$\frac{x^2-<!--\; -\; -->2xy+y^2-<!--\; -\; -->z^2}{x^2+2xy+y^2-<!--\; -\; -->z^2}÷<!--\; ÷\; -->\frac{x-<!--\; -\; -->y+z}{x+y-<!--\; -\; -->z}$$
Then, multiplying the dividend by the reciprocal of the divisor and rewriting.
$$\frac{x^2-<!--\; -\; -->2xy+y^2-<!--\; -\; -->z^2}{x^2+2xy+y^2-<!--\; -\; -->z^2}\times <!--\; \times \; -->\frac{x+y-<!--\; -\; -->z}{x-<!--\; -\; -->y+z}$$
$$\frac{(x^2-<!--\; -\; -->2xy+y^2-<!--\; -\; -->z^2)(x+y-<!--\; -\; -->z)}{(x^2+2xy+y^2-<!--\; -\; -->z^2)(x-<!--\; -\; -->y+z)}$$
Expanding the expression and grouping i got
$$\frac{x^3-<!--\; -\; -->x^2(y+z)-<!--\; -\; -->x(y-<!--\; -\; -->z{)}^2+y^3-<!--\; -\; -->y^{2}z-<!--\; -\; -->y{z}^2+z^3}{x^3+x^2(y+z)-<!--\; -\; -->x(y-<!--\; -\; -->z{)}^2-<!--\; -\; -->y^3+y^{2}z+y{z}^2-<!--\; -\; -->z^3}$$
The problem is that i don't know what to do now. I know that the result is:
$$\frac{x-<!--\; -\; -->y-<!--\; -\; -->z}{x+y+z}$$
I want to belive that there's an easy way from the beggining (to avoid expanding all the thing) but i would like your help.

What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of x.

Transform term for induction proof
Could you help me to mathematically show that these two terms are the same (they are). This is the last (but probably the most important :( ) step of an induction proof.
$$First:\frac{(n+1)(n+2)(2(n+1)+7)}{6}$$
$$Second:\frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}$$
Thank you! :)

How to find the integer values?
Let $N=\frac{3k-8}{k+4}$
where k is an integer.
List all positive values of k for which N is an integer.
Do this using algebra, without using a calculator.
Also do not do this by enumeration, i.e., plugging in $k=1,2,...$ and checking the value of N.

Perform the operation and give the answer as a fraction in lowest term
$$(\frac{9}{10})+(-<!--\; -\; -->\frac{6}{7})$$

Repeated division until the result is greater than a given number
I have a problem of repeated multiplication with a fraction (strictly less than 1), until the result is greater than a given number.
For example: Let's say the given two numbers are 3 and 1.5. We have to multiply 3 repeatedly with 0.667 (say) until the result is greater than 1.5
$$3\cdot <!--\; \cdot \; -->0.667=2.001$$
$$2.001\cdot <!--\; \cdot \; -->0.667=1.33$$ (which is less than 1.5)
So the number of multiplications we made is 1 before we hit a number less than 1.5. Instead of brute-forcing the solution(the number of times we are able to multiply), is there a way that we can find it in a better way?

Finding the range of the rational function $f(x)=\frac{2x-<!--\; -\; -->1}{x+2}$ for $x\in <!--\; \in \; -->[-<!--\; -\; -->1,2]$

Sum on simultaneous equation
Solving this simultaneous equation is little bit hard.
The equation
$$\frac{1}{y-<!--\; -\; -->2}-<!--\; -\; -->\frac{1}{x+2}=\frac{1}{60}$$
and
$$\frac{1}{x}-<!--\; -\; -->\frac{1}{y}=\frac{1}{60}.$$
How to solve it please explain.

What am I doing wrong with this adding fractions problem: $\frac{5}{6}+\frac{2}{3}$?
The problem is
$$\frac{5}{6}+\frac{2}{3}$$
I found the LCD which is 6. Then I multiplied $2\times <!--\; \times \; -->3=6$ and $2\times <!--\; \times \; -->2=4$. So the new problem is
$$\frac{5}{6}+\frac{4}{6}=\frac{11}{6}$$
$\frac{11}{6}$ simplified is $1\frac{5}{6}$ but that's not an answer I can select.

Question about dividing simple fractions
I had some exercises and I came upon something that got me confused.
$$\frac{6\ast <!--\; \ast \; -->{10}^w}{7}$$
Turns out to be
$$\frac{6}{7}\ast <!--\; \ast \; -->\frac{{10}^w}{7}$$
$$\frac{6\ast <!--\; \ast \; -->{10}^w}{49}$$
On the other hand, for example:
$$\frac{12\ast <!--\; \ast \; -->{10}^w}{14}$$
does NOT equal:
$$\frac{2\ast <!--\; \ast \; -->6}{2\ast <!--\; \ast \; -->7}\ast <!--\; \ast \; -->\frac{{10}^w}{7}$$
$$\frac{6}{7}\ast <!--\; \ast \; -->\frac{{10}^w}{7}$$
$$\frac{6\ast <!--\; \ast \; -->{10}^w}{49}$$
Why is that? I'm a bit confused so if I could get cleared up on some rules I've missed, I would appreciate it. Thanks.

Clarification about a proof
it's just a clarification about a proof :
The starting inequality is :
$$0.5\ge <!--\; \ge \; -->1.5-<!--\; -\; -->\frac{1}{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}+1}$$
So we study the function $f(x)=0.5-<!--\; -\; -->\frac{1}{x+1}$ wich is concave so we can apply Jensen inequality we get :
$$0.5-<!--\; -\; -->\frac{1}{\frac{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}}{3}+1}\ge <!--\; \ge \; -->\frac{1}{3}(1.5-<!--\; -\; -->\frac{1}{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}+1})$$
Wich is equivalent to :
$$1.5-<!--\; -\; -->\frac{3}{\frac{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}}{3}+1}\ge <!--\; \ge \; -->(1.5-<!--\; -\; -->\frac{1}{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}+1})$$
But it's easy to find the minimum of :
$$\frac{3}{\frac{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}}{3}+1}$$
Wich is one so we have :
$$0.5\ge <!--\; \ge \; -->1.5-<!--\; -\; -->\frac{1}{\frac{a^2+ab+2a+b^2+3}{a^2+ab-<!--\; -\; -->2a+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{c^2+cb+2b+b^2+3}{c^2+cb-<!--\; -\; -->2b+b^2+3}+1}-<!--\; -\; -->\frac{1}{\frac{a^2+ac+2c+c^2+3}{a^2+ac-<!--\; -\; -->2c+c^2+3}+1}$$
Done !

Find $a,b\in <!--\; \in \; -->\mathbb{N}$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{(a,b)}+\frac{1}{[a,b]}=\frac{1}{2}$
My try :
$$\frac{[a,b]+(a,b)}{[a,b]\cdot <!--\; \cdot \; -->(a,b)}=\frac{ab-<!--\; -\; -->2a-<!--\; -\; -->2b}{2ab}.$$
Now what to do?

Multiplying by Inverse of a Fraction question
Hey so I'm doing an exercise and I got a bit confused by something.
I've learned early on that you can multiply the inverse of a fraction and would get the same result as you would if you divided, since you do the exact opposite.
Then why doesn't this hold true for:
$$\frac{10z^{1/3}}{2z^2}{}^{}=10z^{1/3}\ast <!--\; \ast \; -->2z^{-<!--\; -\; -->2}$$
Please explain in simple words.. not too math savvy ^^

If we know range of a function , how can construct range of other function?
If we know $\frac{-<!--\; -\; -->1}{2}\le <!--\; \le \; -->\frac{x}{x^2+1}\le <!--\; \le \; -->\frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-<!--\; -\; -->1}{2}\le <!--\; \le \; -->\frac{x}{x^2+1}\le <!--\; \le \; -->\frac{1}{2}$

Is division the same thing as a fraction?
So, really basic mathematics question, but would the fraction $2/2$ be the exact same as dividing $2$ by $2$? Would this apply to all fractions and all division problems? Or is it just coincidence? I'm not entirely sure and I get vague answers from my mathematics teacher, so I'm not $100$%.
I figured I'd ask this community to see what everyone thought.

Find all the complex numbers $z$ such that $w=\frac{2z-<!--\; -\; -->1}{z-<!--\; -\; -->2}$ has argument $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$

Relationships between denominator constants from fractional decomposition
Performing fractional decomposition on the fraction
$\frac{a(x-<!--\; -\; -->c)+b}{(x-<!--\; -\; -->y_1)(x-<!--\; -\; -->y_2)}=\frac{A}{x-<!--\; -\; -->y_1}+\frac{B}{x-<!--\; -\; -->y_2}$(1)
where x is a variable and the rest are constants, you find
$A=\frac{a(y_1-<!--\; -\; -->c)+b}{y_1-<!--\; -\; -->y_2}$
$B=\frac{a(y_2-<!--\; -\; -->c)+b}{y_2-<!--\; -\; -->y_1}$
If you try to check these values by inserting A and B into equation (1), you arrive at this equation
$\frac{a(x-<!--\; -\; -->c)+b}{(x-<!--\; -\; -->y_1)(x-<!--\; -\; -->y_2)}=\frac{(y_1-<!--\; -\; -->y_2{)}^2[a(x-<!--\; -\; -->c)+b]}{(x-<!--\; -\; -->y_1)(x-<!--\; -\; -->y_2)}$
which implies that $(y_1-<!--\; -\; -->y_2{)}^2=1$, despite these being just arbitrary constants. Is this true? Where does this come from? I came upon this from checking the values obtained from fractional decomposition for a dynamic model on a first order system.

Solve:
$$\frac{(-<!--\; -\; -->1)+4)}{((-<!--\; -\; -->1)+(-<!--\; -\; -->2))\times <!--\; \times \; -->(-<!--\; -\; -->9)}$$

Can the inequality $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge <!--\; \ge \; -->6$ be proved with differentiation?
with a,b,c>0
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.

Fraction of two prime numbers
Is there any name or special mathematical importance for fractions of two prime numbers, e.g. $$\frac{7}{11}$$ or $$\frac{103}{277}$$?