Master Fraction Math Problems with Expert Help

Partial fractions - alternative result
I have the following fraction:
$\frac{z}{(z - 1) (z - 2)}$
When I try to decompose it to partial fractions, I get:
$- \frac{1}{z - 1} + \frac{2}{z - 2}$
But the result in my book is:
$- \frac{z}{z - 1} + \frac{z}{z - 2}$
Both results are correct, but, how am I supposed to get the second one?

Zion Wheeler 2022-06-07

What exactly goes raising a number to a fraction mean?
I apologise for asking something so fundamental, but what exactly does
$2^{\frac{2}{5}}$
actually mean? I get raising a whole number to another whole number
$x^{y}$
means you are multiplying x with itself y times, but what does it mean when y is a fraction?

Yahir Crane 2022-06-07

Reduce a fraction
Given the function
$f (x) = \frac{x^{2} - 5}{x + \sqrt{5}}$
If I draw this function in maple, I will get a line. How can that be true? I should expect a line except in area of $x = - \sqrt{5}$ , where $f (x) \to \infty$ or $f (x) \to - \infty$ . Of course Maple has factorized the numerator and reduced.
$\frac{x^{2} - 5}{x + \sqrt{5}} = x - \sqrt{5}, x \neq - \sqrt{5}$
My question is now, how can we ever reduce such a fraction with only condition $x \neq - \sqrt{5}$ , when it is " $- \sqrt{5}$ and around it".

Arraryeldergox2 2022-06-07

Summing reciprocal logs of different bases
I recently took a math test that had the following problem:
$\frac{1}{\log_{2} 50!} + \frac{1}{\log_{3} 50!} + \frac{1}{\log_{4} 50!} + \dots + \frac{1}{\log_{50} 50!}$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$\frac{1}{\log_{2} 1 + \log_{2} 2 + \log_{2} 3 + \dots + \log_{2} 50}$
How do the fractions with such irrational values become $1$ ? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?

Tristian Velazquez 2022-06-07

Hard question simplify as much as possible following
$\frac{x^{3} - 49 x}{x^{4} - 13 x^{3} + 50 x^{2} - 56 x}$
I get answer $\frac{x + 3}{x - 1}$ but my friend say it's wrong I dont understand please help me please

gvaldytist 2022-06-07

Intuitively, why does $\frac{a}{c} = \frac{1}{\frac{c}{a}}$ ?
For intuition, I reference objects. Imagine making a dessert with: $a$ as apples and $c$ as chestnuts.
Question. How and why is
$\frac{a}{c} (3) = \frac{1}{\frac{c}{a}} (4) ?$
I ask only about intuition; please omit formal arguments and proofs (eg: Intuition is not generated by the explanation that rationalising (4)'s denominator produces (3)).

Karina Trujillo 2022-06-06

rational numbers as upper limit of a summation?
a quick question: Is it a legit way to use a fraction as the upper limit of a summation? Given is a frequency $f$ and a sample rate $f_{s}$ . I want to use a sum like this:
$\sum_{k = 1}^{\frac{f_{s}}{2 f}} \sin (2 \cdot π \cdot k)$
Is this a legit way to write it?

Leonel Contreras 2022-06-06

Steps to simplify specific fraction
What are the steps to simplify the first fraction to the second?
$\frac{\frac{3 - (3 + h)}{3 (3 + h)}}{h} ⟹ \frac{- 1}{3 (3 + h)}$
I assume the first step would be to re-write it like so:
$\frac{3 - (3 + h)}{3 (3 + h) h}$
But after that I don't know where to go...

taghdh9 2022-06-05

Inequality on non-negative reals
For non-negative $x, y, z$ satisfy $\frac{1}{2 x^{2} + 1} + \frac{1}{2 y^{2} + 1} + \frac{1}{2 z^{2} + 1} = 1$ then show that Idea how to handle the constraint? I'm unaware .

tr2os8x 2022-06-05

What is $\frac{x^{10} + x^{8} + x^{2} + 1}{x^{10} + x^{6} + x^{4} + 1}$ given $x^{2} + x - 1 = 0$ ?

Zion Wheeler 2022-06-05

Simple algebraic manipulation with 2 equations
My first equations is this: $d_{2} = d - 30.$
My second equations is this: $\frac{1}{d_{2}} = \frac{1}{12} - \frac{1}{1.066 (d - 30)}$
I am trying to solve for $d_{2}$ in the second equation and then set the two equations equal to each other to solve for d.
I am having trouble solving for $d_{2}$ in the second equation. I think the fractions are giving me a hard time. Any help is much appreciated, thanks!

Brunton39 2022-06-05

Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801?
I've seen this happen that when you divide in a fraction using the square of any number with only nines in the denominator. Like in
$\frac{1}{9^{2}} = \frac{1}{81} = 0.01234567 \underset{↑}{} 901234567 \underset{↑}{} 901234567 \dots,$
and in
$\frac{1}{99^{2}} = \frac{1}{9801} = 0.0001020304050607080910111213 \dots 9697 \underset{↑}{} 99000102 \dots,$
the decimals go on predictably when suddenly in the first one you miss 8, and in the second you miss 98 and it keeps going on forever. How does this happen? Why do we miss numbers like 8 in the decimal representation of $\frac{1}{9^{2}},$ , or like 98 in the decimal of $\frac{1}{99^{2}},$ or 998 in $\frac{1}{999^{2}},$ , or $9998$ in $\frac{1}{9999^{2}}$ ?

immogiaveoe3h9 2022-06-04

Property of fractions
Given two fractions $\frac{h}{k}$ and $\frac{h^{^{'}}}{k^{^{'}}}$ both in reduced form. I am unable to find a case when $\frac{h + h^{^{'}}}{k + k^{^{'}}}$ does not lie in the interval $[\frac{h}{k}, \frac{h^{^{'}}}{k^{^{'}}}]$ . Is there such a case ?
PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.

Kasen Keller 2022-06-04

Laurent expansion of $\frac{1}{z^{2}}$
I need to find a Laurent expansion of $\frac{1}{z^{2}}$ with centre in $z_{0} = 1$ and $P (1, 2014, 2015)$
If it was $\frac{1}{z}$ , I'd rewrite the fraction like this:
$\frac{1}{(z - 1) + 1}$
But with $z^{2}$ I'm not sure how to proceed...
My attempt:
According to the given recommendation, I've tried to do the expansion of $\frac{1}{z}$ :
$\frac{1}{z} = \frac{1}{1 + (z - 1)}, a_{0} = 1, q = - (z - 1) = 1 - z$
$\sum_{n = 0}^{\infty} (1 - z)^{n} = 1 + (1 - z) + (1 - z)^{2} + . . .$
And after differentiation:
$0 + 1 + 2 (1 - z) + 3 (1 - z)^{2} + . . . = \sum_{n = 0}^{\infty} n (1 - z)^{n - 1}$

Angeline Henderson 2022-06-03

If $\frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c} = 1$ , prove $\sqrt{a b} + \sqrt{a c} + \sqrt{b c} \leq 3$

Alani Patrick 2022-06-03

Conditions present after solving for x
The question asked me to solve
$\frac{x^{2} + 2 x - 8}{x^{2} - x - 2} = 3$
The answer is $x = 0.5$ , which I worked out, but in the answers it says that I must state the condition that $x \neq 2$ to get full marks. Why is this condition present? How can I recognise when a condition exists, when I encounter similar problems?

Emilio Waters 2022-06-03

Decompose a fraction in a sum of two
Let's say that I have this fraction:
$\frac{2 x}{x^{2} + 4 x + 3}$
I would like to decompose in two fraction:
$\frac{A}{x + 3} + \frac{B}{x + 1}$
Which is the procedure for that? :)

Tananiajtac2 2022-06-03

Advanced GCSE Indices
At school, we've been asked to simplify the following:
${(\frac{S^{9 / 10}}{S^{3 / 5}})}^{- 4}$
I've given it a go but haven't been taught this yet. Could someone please explain the process which one would use?

Cameron Pearson 2022-06-03

Find the minimum of $\sum_{c y c} \sqrt{\frac{a}{2 (b + c)}}$ with $a, b, c > 0$
As said in the title, I have to find the minimum of the following:
$\sum_{c y c} \sqrt{\frac{a}{2 (b + c)}}$
with $a + b + c > 0$
In my very last attempt, I tried to work it out using AM-GM: Since
$\sqrt{\frac{a}{2 (b + c)}} = \frac{1}{\sqrt{2}} \frac{a}{\sqrt{a (b + c)}} \geq \frac{\sqrt{2} a}{a + b + c}$
$\Rightarrow \sum_{c y c} \sqrt{\frac{a}{2 (b + c)}} \geq \frac{\sqrt{2} (a + b + c)}{a + b + c} = \sqrt{2}$
The problem is, to get the minimum value,
$a = b + c$
$b = c + a$
$c + a + b$
or $a = b = c = 0$ . This is wrong since $a, b, c > 0$ And after some calculus I found the minimum value is $\frac{3}{2}$ when $a = b = c$ . So what mistake I have made?

Sturmboot0ae2a 2022-06-03

Adding 1 to a/b?
I've been learning algebra $2$ and came across a question telling me to add $1$ to $\frac{a}{b}$ by finding the least common multiple of the divisors. So I got $\frac{1 b}{1 b} + \frac{a}{b}$ that simplified is $\frac{b}{b} + \frac{a}{b}$ . So then I added those two fractions together and got $\frac{a b}{b}$ I saw that the two $b$ 's cancelled each other out so I got $\frac{a}{1}$ which is $a$ ? How does adding $1$ to $\frac{a}{b}$ get $a$ , this doesn't seem to be right because $1 + \frac{2}{3}$ isn't $2$ it's $1$ and $\frac{2}{3}$ . Am I doing something wrong or is this actually correct, if so can you explain to me why?

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