Question

спростити вираз 1/(a*(a+3))+1/((a+3)*(a+6))+1/((a+6)*(a+9))+1/((a+9)*(a+12))+1/((a+12)*(a+15))

Answer 1

Смотри...................

Answer 2

Ответ:   $\frac{5}{a(a+15)}$ .

Объяснение:

$\frac{1}{a(a+3)}=\frac{1/3}{a}-\frac{1/3}{a+3}\\\\\frac{1}{(a+3)(a+6)}=\frac{1/3}{a+3}-\frac{1/3}{a+6}\\\\\frac{1}{(a+6)(a+9)}=\frac{1/3}{a+6}-\frac{1/3}{a+9}\\\\\frac{1}{(a+9)(a+12)}=\frac{1/3}{a+9}-\frac{1/3}{a+12}\\\\\frac{1}{(a+12)(a+15)}=\frac{1/3}{a+12}-\frac{1/3}{a+15}\\\\\\\frac{1}{a(a+3)}+\frac{1}{(a+3)(a+6)}+\frac{1}{(a+6)(a+9)}+\frac{1}{(a+9)(a+12)}+\frac{1}{(a+12)(a+15)}=\\\\=\frac{1/3}{a}-\frac{1/3}{a+3}+\frac{1/3}{a+3}-\frac{1/3}{a+6}+\frac{1/3}{a+6}-\frac{1/3}{a+9}+\frac{1/3}{a+9}-\frac{1/3}{a+12}+\frac{1/3}{a+12}-\frac{1/3}{a+15}=\\\\=\frac{1/3}{a}-\frac{1/3}{a+15}=\frac{1}{3}\cdot \Big (\frac{1}{a}-\frac{1}{a+15}\Big )=\frac{1}{3}\cdot \frac{a+15-a}{a(a+15)}=\frac{1}{3}\cdot \frac{15}{a(a+15)}=\frac{5}{a(a+15)}$