Simplification: (3x^2)/(1-x^2)^(3/2) + 1/(1-x)^(5/2) = (2x^2+1)/(1-x^2)^(5/2)

[bluemath]

Hello,

How to proove this equality please ?

My result is the first two terms. But my software proposes a simplication and I don't understand how it finds that (final result)

\(\displaystyle \frac{3x^{2}}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{1}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\)

Thanks

Edit : (sorry, I did a misstake)

\(\displaystyle \frac{1}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{3x^{2}}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\) for all x /{1}

 

[Ishuda]

Hello,

How to proove this equality please ?

My result is the first two terms. But my software proposes a simplication and I don't understand how it finds that (final result)

\(\displaystyle \frac{3x^{2}}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{1}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\)

Thanks

Plug in x=1/2 and check to see if that really is an equality for all x (which is what I assume you meant - but watch out for that word ***-u-me)

 

[bluemath]

Edit : (sorry, I did a misstake)

\(\displaystyle \frac{1}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{3x^{2}}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\) for all x /{1}

 

[bluemath]

I found the error, saddened for the disturbance.

 

[ksdhart]

Wait, hold on. Neither your original equality, nor the edited version are actual equalities, for all values of x. The edited version:

\(\displaystyle \displaystyle \frac{1}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{3x^2}{\left(1-x\right)^{\frac{5}{2}}}=\frac{2x^2+1}{\left(1-x^2\right)^{\frac{5}{2}}}\)

Has only one real number solution, when x = 0. The original version, meanwhile:

\(\displaystyle \displaystyle \frac{3x^2}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{\left(1-x\right)^{\frac{5}{2}}}=\frac{2x^2+1}{\left(1-x^2\right)^{\frac{5}{2}}}\)

Has two real solutions, x = 0 and x ~ 1.30944, according to WolframAlpha.

Is there perhaps yet another typo somewhere? Otherwise, I don't see how this make any sense.

 

[Ishuda]

Hello,

How to proove this equality please ?

My result is the first two terms. But my software proposes a simplication and I don't understand how it finds that (final result)

\(\displaystyle \frac{3x^{2}}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{1}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\)

Thanks

Edit : (sorry, I did a misstake)

\(\displaystyle \frac{1}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{3x^{2}}{({1-x)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\) for all x /{1}

Assuming what you meant was
\(\displaystyle \frac{1}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{3x^{2}}{({1-x^2)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\)
that is, the denominator of the second expression on the left hand side is \(\displaystyle (1-x^2)^{3/2}\) and not \(\displaystyle (1-x)^{3/2}\),
just put the two expressions on the right over a common denominator.

 

[bluemath]

Assuming what you meant was
\(\displaystyle \frac{1}{({1-x^{2})}^{3/2}}\) + \(\displaystyle \frac{3x^{2}}{({1-x^2)}^{5/2}}\) = \(\displaystyle \frac{2x^{2}+1}{({1-x^{2})}^{5/2}}\)
that is, the denominator of the second expression on the left hand side is \(\displaystyle (1-x^2)^{3/2}\) and not \(\displaystyle (1-x)^{3/2}\),
just put the two expressions on the right over a common denominator.

Exactly, thank you Ishuda