Ответ:

[tex]\dfrac{a-2}{a^2-4} -\dfrac{a-2}{4-a^2} =\dfrac{2}{a+2}[/tex]

[tex]\dfrac{15x-2}{10x^2} +\dfrac{5+x}{5x^3} =\dfrac{3x^2+2}{2x^3}[/tex]

[tex]\dfrac{7}{x^2+x}+\dfrac{13}{x+1} =\dfrac{7+13x}{x^2+x}[/tex]

Решение:

1)

[tex]\dfrac{a-2}{a^2-4} -\dfrac{a-2}{4-a^2} =\dfrac{a-2}{a^2-4} +\dfrac{a-2}{a^2-4} =\dfrac{2(a-2)}{a^2-4} =[/tex]

[tex]=\dfrac{2(a-2)}{a^2-2^2} =\dfrac{2(a-2)}{(a-2)(a+2)} =\boxed{\dfrac{2}{a+2}}[/tex]

2)

[tex]\dfrac{15x-2}{10x^2} +\dfrac{5+x}{5x^3} =\dfrac{(15x-2)\cdot x}{10x^3} +\dfrac{(5+x)\cdot2}{10x^3}=[/tex]

[tex]=\dfrac{15x^2-2x}{10x^3} +\dfrac{10+2x}{10x^3} =\dfrac{15x^2-2x+10+2x}{10x^3} =[/tex]

[tex]=\dfrac{15x^2+10}{10x^3} =\dfrac{5(3x^2+2)}{10x^3} =\boxed{\dfrac{3x^2+2}{2x^3}}[/tex]

3)

[tex]\dfrac{7}{x^2+x}+\dfrac{13}{x+1} =\dfrac{7}{x^2+x}+\dfrac{13x}{x(x+1)} =[/tex]

[tex]=\dfrac{7}{x^2+x}+\dfrac{13x}{x^2+x} =\boxed{\dfrac{7+13x}{x^2+x} }[/tex]