Can someone please explain how to solve these 3 problems? It says “simplify the following fractions”

When adding fractions, you must express them over a common denominator. The easiest common denominator to find is the product of the denominators of the individual fractions. That is ...
[tex]\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}[/tex]
When dividing fractions, multiply by the inverse of the denominator fraction.
[tex]\displaystyle\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}[/tex]
12.
[tex]\displaystyle\frac{\frac{2}{x}-3}{1-\frac{1}{x-1}}=\frac{\frac{2-3x}{x}}{\frac{x-1-1}{x-1}}\\\\=\frac{(2-3x)(x-1)}{x(x-2)}=\frac{-3x^2+5x-2}{x^2-2x}[/tex]
13.
[tex]\displaystyle\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x}+\frac{1}{y}}=\frac{\frac{y^2-x^2}{x^{2}y^{2}}}{\frac{y+x}{xy}}\\\\=\frac{\left(y^2-x^2\right)xy}{(y+x)x^{2}y^{2}}=\frac{(y-x)(y+x)}{(y+x)xy}=\frac{y-x}{xy}[/tex]
14.
[tex]\displaystyle\frac{\frac{x}{2}-1}{x-2}=\frac{\frac{x-2}{2}}{x-2}=\frac{x-2}{2}\cdot\frac{1}{x-2}=\frac{1}{2}[/tex]