Melody

Thanks for joining in anon.  What are the rules for your puzzle.   

8, 3, 4, 7

The magic number is 24. Good luck!

3*8=24

Do I need to use all the numbers or what?

That is a great answer asinus but I suspect it was not the intended question :)

$$\\-11x^2 + 2x = 10\\\\
-11x^2 + 2x -10 = 0\\\\
11x^2 - 2x +10 = 0\\\\
\triangle = 4-4*11*10 <0\\\\
$This has no real roots because the discriminant is less than zero.$\\$It does have imaginary roots but I suspect you are not interested in those. :)$$$

Come on Denzel it is not the end of the world :)

If you want to improve your maths why don't you post more questions. 

They can be general questions if you like.  We can send you off to look at video clips or we can help you directly.  :)

Anyway, like I said, it is not the end of the world.  So cheer up :)

$${\frac{\left({\frac{{\mathtt{x}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{y}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{{\mathtt{y}}}^{{\mathtt{4}}}\right)}}\right)}{\left({\frac{{\mathtt{x}}}{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}\left({{\mathtt{y}}}^{{\mathtt{2}}}\right)\right)}}\right)}}$$

$$\\\left(\left(\frac{x}{(x+y)}+\frac{y}{(x-y)}\right)*\left(\frac{(x^2-xy)}{(x^4-y^4)}\right)\right)\div\left(\frac{x}{(x^2)+2xy+(y^2)}\right)\\\\
=\left(\left(\frac{x(x-y)}{(x+y)(x-y)}+\frac{y(x+y)}{(x-y)(x+y)}\right)*\left(\frac{x(x-y)}{(x^2)^2-(y^2)^2}\right)\right)\times\left(\frac{(x^2)+2xy+(y^2)}{x}\right)\\\\
=\left(\left(\frac{x^2-xy+xy+y^2}{(x+y)(x-y)}\right)*\left(\frac{x(x-y)}{(x^2)^2-(y^2)^2}\right)\right)\times\left(\frac{(x+y)^2}{x}\right)\\\\
=\left(\left(\frac{x^2+y^2}{(x+y)}\right)*\left(\frac{x}{(x^2-y^2)(x^2+y^2)}\right)\right)\times\left(\frac{(x+y)^2}{x}\right)\\\\
=\frac{x^2+y^2}{1}\times\frac{1}{(x^2-y^2)(x^2+y^2)}\times\frac{x+y}{1}\\\\
=\frac{1}{1}\times\frac{1}{(x^2-y^2)}\times\frac{x+y}{1}\\\\
=\frac{1}{(x-y)(x+y)}\times\frac{x+y}{1}\\\\
=\frac{1}{x-y}\\\\$$

2.

(1/(x-2) - 2/(x+1) + 3/x) * (((x^2)+x)/(x^2+x-3)))

$$\left({\frac{{\mathtt{1}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{x}}}}\right){\mathtt{\,\times\,}}\left({\frac{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}\right)$$

$$\\\left(\frac{1}{(x-2)} -\frac{ 2}{(x+1)} + \frac{3}{x}\right) \times \frac{x(x+1)}{x^2+x-3}\\\\
=\left(\frac{x(x+1)}{x(x+1)(x-2)} -\frac{ 2x(x-2)}{x(x-2)(x+1)} + \frac{3(x+1)(x-2)}{x(x+1)(x-2)}\right) \times \frac{x(x+1)}{x^2+x-3}\\\\
=\frac{x(x+1)-2x(x-2)+3(x+1)(x-2)}{x(x+1)(x-2)} \times \frac{x(x+1)}{x^2+x-3}\\\\
=\frac{x^2+x-2x^2+4x+3(x^2-x-2)}{(x-2)} \times \frac{1}{x^2+x-3}\\\\
=\frac{x^2+x-2x^2+4x+3x^2-3x-6}{(x-2)} \times \frac{1}{x^2+x-3}\\\\
=\frac{2x^2+2x-6}{(x-2)} \times \frac{1}{x^2+x-3}\\\\
=\frac{2(x^2+x-3)}{(x-2)} \times \frac{1}{x^2+x-3}\\\\
=\frac{2}{x-2} \\\\$$

Again, you need to check my working although this one looks more promising than the first one that I did :)

1.     ((x/(x^2-1)) - (3/(x+1)))/(2x^2-x-3)/(x^3-1)

YOU WILL NEED TO CHECK MY ANSWER, I COULD EASILY HAVE MADE ONE OR MORE CARELESS ERRORS.

$$\\\left(\frac{x}{(x^2-1)} - \frac{3}{(x+1)}\right)\div \frac{(2x^2-x-3)}{(x^3-1)}\\\\
= \left(\frac{x}{(x-1)(x+1)} - \frac{3}{(x+1)}\right)\div \frac{(2x^2+2x-3x-3)}{(x-1)(x^2+x+1)}\\\\
= \left(\frac{x}{(x-1)(x+1)} - \frac{3(x-1)}{(x+1)(x-1)}\right)\div \frac{2x(x+1)-3(x+1))}{(x-1)(x^2+x+1)}\\\\
= \left(\frac{x}{(x-1)(x+1)} - \frac{3x-3}{(x+1)(x-1)}\right)\div \frac{(2x-3)(x+1)}{(x-1)(x^2+x+1)}\\\\
= \frac{x-3x+3}{(x-1)(x+1)} \times \frac{(x-1)(x^2+x+1)}{(2x-3)(x+1)}\\\\
= \frac{-2x+3}{(x+1)} \times \frac{(x^2+x+1)}{(2x-3)(x+1)}\\\\
= \frac{-1(2x-3)}{(x+1)} \times \frac{(x^2+x+1)}{(2x-3)(x+1)}\\\\
= \frac{-1}{(x+1)} \times \frac{(x^2+x+1)}{(x+1)}\\\\
=-\frac{x^2+x+1}{x^2+2x+1}\\\\$$

It depends on where your brackets were meant to be :/

$$\\\frac{log (13) x^7 \sqrt{y}}{z^4}\\\\
OR\\\\
=log (13) x^7 \sqrt{\frac{y}{z^4}}\\\\
=log (13) x^7 \frac{\sqrt{y}}{z^2}\\\\
=\frac{x^7\sqrt{y\;}\;log (13)}{z^2}$$

Good one TR     

Marcus helped fill water bottles for athletes during the school sports meeting.

The capacity of a bottle was  $${\frac{{\mathtt{4}}}{{\mathtt{13}}}}$$    litres.

Does 7/8 divided by 7/10 equal 1 and 1/4?

let's  see  

$$\\\frac{7}{8}\div \frac{7}{10}\\\\
=\frac{\not{7}}{\not{8}^4}\times \frac{\not{10}^5}{\not{7}}\\\\
=\frac{5}{4}\\\\
=1\frac{1}{4}$$

So yes it does :)