LaTeX Where can I test LaTex without disrupting discussions?

[Hollysmoke]

x^2-6x-7/2x^2+3x +1

 

[Hollysmoke]

\frac{x^2-6x-7}{2x^2+3x+1}

 

[Hollysmoke]

\frac{x^2-6x-7}{2x^2+3x+1} x \frac{x^2-6x-7}{2x^2+3x+1} / \frac{x^2-6x-7}{2x^2+3x+1}

 

[Hollysmoke]

\frac{x^2-6x-7}{2x^2+3x+1} x \frac{4x^2-16x+15}{2x^2-17x+21} / \frac{4x^2-20x+25}{4x^2-1}

 

[Hollysmoke]

\frac{(x-7)(x+1)}{(2x+1)(x+1)} x \frac{(2x-5)(2x-3)}{2x-3)(x-7)} x \frac{2x+1)(2x-1)}{(2x-5)(2x+5)}

 

[Hollysmoke]

\frac{(x-7)(x+1)}{(2x+1)(x+1)} x \frac{(2x-5)(2x-3)}{(2x-3)(x-7)} x \frac{2x+1)(2x-1)}{(2x-5)(2x+5)}

 

[Hollysmoke]

\frac{(x-7)(x+1)}{(2x+1)(x+1)} x \frac{(2x-5)(2x-3)}{2x-3)(x-7)} x \frac{(2x+1)(2x-1)}{(2x-5)(2x+5)}

 

[Hollysmoke]

\frac{(x-7)(x+1)}{(2x+1)(x+1)} x \frac{(2x-5)(2x-3)}{(2x-3)(x-7)} x \frac{(2x+1)(2x-1)}{(2x-5)(2x+5)}

 

[Hollysmoke]

\frac{(2x-1)}{(2x+5)}

 

[Hollysmoke]

\frac{5}{(x+3)} - \frac{(1)}{x-1}

 

[Hollysmoke]

\frac{5}{(x+3)} - \frac{1}{x-1}

 

[Hollysmoke]

\frac{5}{(x+3)} - \frac{(1)}{(x-1)}
\frac{5(x-1)}{(x+3)(x-1)} - \frac{(x+3))}{(x-1)(x+3)}
\frac{5(x-1)-(x+3)}{(x+3)(x-1)}
\frac{5x-5-x-3}{(x+3)(x-1)}
\frac{4x-8}{(x+3)(x-1)}
\frac{4(x-2)}{(x+3)(x-1)}

 

[NateTG]

\frac{x^2-6x-7}{2x^2+3x+1} x \frac{4x^2-16x+15}{2x^2-17x+21} / \frac{4x^2-20x+25}{4x^2-1}

Did you mean:
\frac{x^2-6x-7}{2x^2+3x+1} \times \frac{\frac{4x^2-16x+15}{2x^2-17x+21}}{\frac{4x^2-20x+25}{4x^2-1}}

 

[Doc Al]

Hollysmoke:
Realize that we have a thread dedicated just to testing Latex: https://www.physicsforums.com/showthread.php?t=8997

Please do your experimenting there and not here.