Melody

$$1/2x^{-1/2} *((2xe^x)/ln(3x^2 +2))\\\\
=\frac{x^{-1/2}}{\not{2}}*\frac{\not{2}xe^x}{ln(3x^2 +2)}\\\\
=\frac{1}{1}*\frac{x^{(1-1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{x^{(1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{\sqrt{x}e^x}{ln(3x^2 +2)}\\\\$$

I don't think that there is anything else to be done with this.  

Has anyone got any further/different thoughts?

$$1/3(8y-2)+1/9(12y+10)\\\\
=\frac{(8y-2)}{3}+\frac{12y+10}{9}\\\\
=\frac{(8y-2)*3}{3*3}+\frac{12y+10}{9}$$

Can you take it from here?

$$\left(\dfrac{-2a^4}{b}\right)^2\\\\\\
=\dfrac{(-2)^2(a^4)^2}{b^2}\\\\
=\dfrac{4a^8}{b^2}\\\\$$

$$(x+4)(x+4)\\
=x(x+4)+4(x+4)\\
=x*x+x*4 \quad +4*x+4*4\\
=x^2+4x+4x+16\\
=x^2+8x+16$$

**I do not know which side has the even or odd, only that both exist.**

Huh?  I am afraid you talk in riddles David. 

Any square number has to have an odd number of factors because all factors come in pairs but the squared one is doubled.  Multiply a square number by 5 and it has to have an odd number plus one more. That is it has to have an even number of factors.

Isn't what I said right?   

Monday 1/9/14

1) This one looks interesting.

@@ End of Day Wrap:    Sun 31/8/14       Sydney, Australia      Time 1:00am (Really Monday morning)     ♬

Hi all,

Our great answerers today included Honga, DavidQD, AzizHusain, Will85237 and CPhill.  Thanks all.  

I'd like to extend a very big welcome to Honga who is new to our web2.0calc forum.  We hope you really like it here.  

Now for a few interesting posts:

* 1) Proove that $${\sqrt{{\mathtt{5}}}}$$ is irrational.

I don't understand the question.  

Just use the site calc.  It is on the home page.  

I only just realised that Honga had answered this.  thanks Honga.

Alan answered a similar question only a couple of days ago.  This is the thread.