Where can I test LaTex without disrupting discussions?

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[tex]x^2-6x-7/2x^2+3x +1[/tex]
 
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[tex]\frac{x^2-6x-7}{2x^2+3x+1}[/tex]
 
[tex]\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}[/tex]
 
[tex]\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}[/tex]
 
[tex]\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)}[/tex]
 
[tex]\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)}[/tex]
 
[tex]\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)}[/tex]
 
[tex]\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)}[/tex]
 
[tex]\frac{(2x-1)}{(2x+5)}[/tex]
 
[tex]\frac{5}{(x+3)} - \frac{(1)}{x-1}[/tex]
 
[tex]\frac{5}{(x+3)} - \frac{1}{x-1}[/tex]
 
[tex]\frac{5}{(x+3)} - \frac{(1)}{(x-1)}[/tex]
[tex]\frac{5(x-1)}{(x+3)(x-1)} - \frac{(x+3))}{(x-1)(x+3)}[/tex]
[tex]\frac{5(x-1)-(x+3)}{(x+3)(x-1)}[/tex]
[tex]\frac{5x-5-x-3}{(x+3)(x-1)}[/tex]
[tex]\frac{4x-8}{(x+3)(x-1)}[/tex]
[tex]\frac{4(x-2)}{(x+3)(x-1)}[/tex]
 
Hollysmoke said:
[tex]\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}[/tex]

Did you mean:
[tex]\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}}[/tex]
 
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