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Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):

$$ \frac{ (a^{1/m}-a^{1/n})^{2} \cdot 4a^{(m+n)/mn} }{ (a^{2/m}-a^{2/n}) (\sqrt[m]{a^{m+1}} + \sqrt[n]{a^{n+1}}) }$$

Here's my try at it: $$\frac{ (a^{1/m} - a^{1/n}) (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} }{ (a^{1/m} - a^{1/n}) (a^{1/m} + a^{1/n}) \cdot a (a^{1/m} + a^{1/n}) }$$

...which is

$$\frac{ (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} }{ a (a^{1/m} + a^{1/n})^2 }$$

Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:

$$\frac{ 1 }{ a (a^{1/m} - a^{1/n}) }$$

How did they do it?

PS I agree with @You're In My Eye that there's a misprint and instead of multiplication in the numerator there should be an addition sign. I want to express my sincere gratitude to everyone who spent their time to help me. Thank you guys very much.

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  • $\begingroup$ I forgot to use parentheses in the original expression, sorry. Fixed it. $\endgroup$
    – Kinkou
    Jul 28, 2016 at 9:41
  • $\begingroup$ Just to check - there is no conditions/restrictions on $a,m,n$ ? $\endgroup$
    – Yuriy S
    Jul 28, 2016 at 9:42
  • $\begingroup$ No, none at all :/ $\endgroup$
    – Kinkou
    Jul 28, 2016 at 9:48
  • $\begingroup$ @kinkou Yes there is some constrain for a and m and n see this page: counter example for this simplification . $\endgroup$
    – Zau
    Jul 28, 2016 at 9:59
  • $\begingroup$ @Kinkou, I found the book but the problem 1.029 is different there: $$\frac{(x^{2/m}-9x^{2/n})(x^{1/m-1}-3x^{1/n-1})}{(x^{1/m}+3x^{1/n})^2-12x^{1/m+1/n}}$$ Looks close, but note the minus sign where you have a multiplication $\endgroup$
    – Yuriy S
    Jul 28, 2016 at 13:50

4 Answers 4

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After checking the older edition of the book, I'm quite sure that the original problem looked like this:

$$\frac{ (a^{1/m}-a^{1/n})^{2} \color{blue}{+} 4a^{(m+n)/mn} }{ (a^{2/m}-a^{2/n}) (\sqrt[m]{a^{m+1}} + \sqrt[n]{a^{n+1}}) }$$

Now we have:

$$(a^{1/m}-a^{1/n})^{2} \color{blue}{+} 4a^{(m+n)/mn} =(a^{1/m}+a^{1/n})^{2}$$

$$(a^{2/m}-a^{2/n}) (\sqrt[m]{a^{m+1}} + \sqrt[n]{a^{n+1}})=a(a^{1/m}-a^{1/n})^{2}(a^{1/m}+a^{1/n})^2$$

Finally we get:

$$\frac{ (a^{1/m}-a^{1/n})^{2} \color{blue}{+} 4a^{(m+n)/mn} }{ (a^{2/m}-a^{2/n}) (\sqrt[m]{a^{m+1}} + \sqrt[n]{a^{n+1}}) }=\frac{1}{a(a^{1/m}-a^{1/n})}$$

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  • $\begingroup$ Seems like this is it! My copy of the book is a 2011 edition, and in this particular problem there's the multiplication sign. I triple checked everything before posting a question here. $\endgroup$
    – Kinkou
    Jul 29, 2016 at 9:57
  • $\begingroup$ @Kinkou, too bad. Skanavi is a classic $\endgroup$
    – Yuriy S
    Jul 29, 2016 at 10:06
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This is not an answer but a counterexample for the undefined constrain.

It cannot be simplified as form $\frac{ 1 }{ a (a^{1/m} - a^{1/n}) }$.

Let $a =-1,m=-1,n=-1$

, where $$\frac{ (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} }{ a (a^{1/m} + a^{1/n})^2 } =\frac{ ((-1)^{1/(-1)} - (-1)^{1/(-1)}) \cdot 4 (-1)^{(1/(-1)) + (1/(-1))} }{ (-1) ((-1)^{1/(-1)} + (-1)^{1/(-1)})^2 } = 0 $$

But $$\frac{ 1 }{ a (a^{1/m} - a^{1/n}) } = undefined $$

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$$\frac{ (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} }{ a (a^{1/m} + a^{1/n})^2 } = \\ \frac{ (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} (a^{1/m} - a^{1/n}) }{ a (a^{1/m} + a^{1/n})^2 (a^{1/m} - a^{1/n}) }$$

So one need simplify: $$\\ \frac{ (a^{1/m} - a^{1/n})^2 \cdot 4 a^{(1/m) + (1/n)} }{ (a^{1/m} + a^{1/n})^2 } = \frac{4(a^{2/m} +a^{2/n} - 2 a^{1/m+1/n})a^{(1/m) + (1/n)}}{{ (a^{1/m} + a^{1/n})^2 } } = \frac{4(a^{3/m+1/n} +a^{3/n+1/m} - 2 a^{2/m+2n})}{{ (a^{1/m} + a^{1/n})^2 } }$$

This is the nearest simplification that I could reach. It cannot be simplified as 1 unless $ 4(a^{3/m+1/n} +a^{3/n+1/m} - 2 a^{2/m+2n}) = {{ (a^{1/m} + a^{1/n})^2 } }$

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  • $\begingroup$ I believe you forgot to include the a term in the denominator in the expression right after '...one need to simplify'. But yeah, same here. $\endgroup$
    – Kinkou
    Jul 28, 2016 at 10:19
  • $\begingroup$ Not yet because I want to simplify those term no in the answer if they come to 1 then the original form should be the answer. $\endgroup$
    – Zau
    Jul 28, 2016 at 10:20
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If you state that:

$$\frac{ (a^{1/m} - a^{1/n}) \cdot 4 a^{(1/m) + (1/n)} }{ a (a^{1/m} + a^{1/n})^2 }=\frac{ 1 }{ a (a^{1/m} - a^{1/n}) }$$

You get a condition on $a,m,n$, which is not satisfied for all values.

Set $p=n/m$, then:

$$2(a^{(3p+1)/2}-a^{(p+3)/2}) = \pm (1+a^p) \tag{1}$$

If you pick values of $a,p$ satisfying this equation, then the last 'simplification' will be correct. But not in general.

We can simplify $(1)$ if we make tow substitutions:

$$a^{1/2}=x,~~~a^{p/2}=y$$

Then $(1)$ becomes:

$$2xy^3 \pm (1+y^2)-2x^3y=0 \tag{2}$$

We can solve $(2)$ as a cubic equation for $x$ or for $y$.

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