Two quite nice problems

As time decides

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Hello people. I got two problems I'd like to solve, but I simply can't. :(

Problem 1 is

\(\displaystyle \displaystyle{\frac {a} {\sqrt{ab} +a} + \frac {b} {\sqrt {ab} - b} - \frac {a} {a - b}}\)

I have worked some with it. One method I tried was to put everything as square roots, but I haven't made any progress.

A pal suggested that variables could neutralize each other straight away. Example: \(\displaystyle \displaystyle{\frac {a} {\sqrt{ab} +a} = \frac {1} {\sqrt{ab}}}\)

Problem 2 is

\(\displaystyle \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}\)

I have only worked a bit on this problem. I think some working with the denominator could help us solve it.

Now these aren't meant to have a solution, just a simpler way of form. Please help me with these two problems :D :D

P.S. Mods: I am new here, and I think this is the right section. Please don't bring forth the BanHammer. :?
 

stapel

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There is no "solving" to do for either of these, since they aren't equations. :shock:

Were the instructions something like "simplify the expression"...? :?:

Thank you! :D

Eliz.

:arrow: P.S. Welcome to FreeMathHelp! 8-)
 

As time decides

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Were the instructions something like "simplify the expression"...?
Yes :)
 

Mrspi

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As time decides said:
Hello people. I got two problems I'd like to solve, but I simply can't. :(

Problem 1 is

\(\displaystyle \displaystyle{\frac {a} {\sqrt{ab} +a} + \frac {b} {\sqrt {ab} - b} - \frac {a} {a - b}}\)

I have worked some with it. One method I tried was to put everything as square roots, but I haven't made any progress.

A pal suggested that variables could neutralize each other straight away. Example: \(\displaystyle \displaystyle{\frac {a} {\sqrt{ab} +a} = \frac {1} {\sqrt{ab}}}\)


P.S. Mods: I am new here, and I think this is the right section. Please don't bring forth the BanHammer. :?
I'd suggest that you eliminate the radical from the denominators of the first two fractions. Multiply numerator and denominator of the first fraction by [sqrt(ab) - a]. Multiply numerator and denominator of the second fraction by [sqrt(ab) + b]. Then, simplify the fractions (you'll have a common factor that divides out in each).

And, I think you will find it fairly easy to get the three fractions to have the same denominator so that they can be combined.

If you still have difficulty with this one, please repost showing all of the work you've done.....
 

Subhotosh Khan

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As time decides said:
Problem 1 is

\(\displaystyle \L \displaystyle{\frac {a} {\sqrt{ab} +a} + \frac {b} {\sqrt {ab} - b} - \frac {a} {a - b}}\)

Let

a/b = x^2

then

a/[sqrt(ab) + a] + b/[sqrt(ab) - b] - a/(a-b)

= 1/[1/x + 1] + 1/[x - 1] - 1/[1-1/x^2]

= x/(x+1) + 1/(x-1) - x^2/(x^2 - 1)

Now continue....


Problem 2 is

\(\displaystyle \L \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}\)

= (25)^2 * 16^{2(n-1)}/[5^4 * 4^{4*(n-2)}]...........edited

Now continue..
 

As time decides

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=)

Thanks for all the replies.

I'd suggest that you eliminate the radical from the denominators of the first two fractions. Multiply numerator and denominator of the first fraction by [sqrt(ab) - a]. Multiply numerator and denominator of the second fraction by [sqrt(ab) + b]. Then, simplify the fractions (you'll have a common factor that divides out in each).
Hm I do not understand what you mean, care to repeat it? (look @ bold sentence)

And, I think you will find it fairly easy to get the three fractions to have the same denominator so that they can be combined.
I thought of a common denominator as a first step but getting a common denominator at that stage takes quite a bit of writing, I advise not to try that approach :wink:

Are you suggesting that we should go for a common denominator as a second step towards a simplification?

Let

a/b = x^2

then

a/[sqrt(ab) + a] + b/[sqrt(ab) - b] - a/(a-b)

= 1/[1/x + 1] + 1/[x - 1] - 1/[1-1/x^2]

= x/(x+1) + 1/(x-1) - x^2/(x^2 - 1)[/tex]

Problem 2 is

\L \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}

= (25) * 16^(n-1)/[5^4 * 4^{4*(n-2)}]

Now continue....
Ooo raw latex code, let's put that in formula.

\(\displaystyle \frac {a} {b} = x^2
\\
then
\\
\frac {a} {sqrt(ab) + a} + \frac {b} {sqrt(ab) - b} - \frac {a} {(a-b)}
\\
= \frac {1} {1/x + 1} + \frac {1} {x - 1} - \frac {1} {1-1/x^2}
\\
= \frac {x} {(x+1)} + \frac {1} {(x-1)} - \frac {x^2} {(x^2 - 1)}\)

Problem 2 is

\(\displaystyle \L \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}
//
= \frac {(25)^2 * 16^{2(n-1)}} {{5^4 * 4^{4*(n-2)}}\)

Thank you very much Subutosh Khan :D

Do you think we could go further with problem nr 1?

Could you take it more in depht how you arrived at

\(\displaystyle \L\displaystyle\frac{(25)^2 * 16^{(n-1)}} {5^4 * 4^{4*(n-2)}}\)
 

Subhotosh Khan

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Re: =)

As time decides said:
Ooo raw latex code, let's put that in formula.

\(\displaystyle \frac {a} {b} = x^2
\\
then
\\
\frac {a} {sqrt(ab) + a} + \frac {b} {sqrt(ab) - b} - \frac {a} {(a-b)}
\\
= \frac {1} {1/x + 1} + \frac {1} {x - 1} - \frac {1} {1-1/x^2}
\\
= \frac {x} {(x+1)} + \frac {1} {(x-1)} - \frac {x^2} {(x^2 - 1)}\)

Problem 2 is

\(\displaystyle \L \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}

= \frac{(25)^2 * 16^{2(n-1)}} {5^4 * 4^{4*(n-2)}}\).........edited

Thank you very much Subutosh Khan :D

Do you think we could go further with problem nr 1?...yes ..you could and should
 

Subhotosh Khan

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Re: =)

As time decides said:
Could you take it more in depht how you arrived at

\(\displaystyle \L\displaystyle\frac{(25)^2 * 16^{2 * (n-1)}} {5^4 * 4^{4 * (n-2)}}\)
9*16^(n-1) + 16^(n)

= 16^(n-1) * 9 + 16 * 16^(n-1)

= 16^(n-1) * [9 + 16]

and

4^(n-1) + 4^(n-2)

= 4*4^(n-2) +1* 4^(n-2)

And continue...
 

Denis

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As time decides said:
A pal suggested that variables could neutralize each other straight away. Example: \(\displaystyle \displaystyle{\frac {a} {\sqrt{ab} +a} = \frac {1} {\sqrt{ab}}}\)
That is VERY incorrect; is 2 / (5 + 2) = 1 / 5 ?

If you are unable to deal with something so basic, you (and your pal!) need classroom felp.
 
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