Đáp án:
Ta có :
$\frac{1}{q}$ + $\frac{1}{q(q+1)}$ = $\frac{q}{q(q+1)}$ + $\frac{1}{q(q+1)}$ = $\frac{q+1}{q(q+1)}$
= $\frac{1}{q}$
b, $\frac{a}{b}$ = $\frac{1}{q+1}$ + $\frac{a(q+1)-b}{b(q+1)}$
= $\frac{b}{b(q+1)}$ + $\frac{a(q+1)-b}{b(q+1)}$
= $\frac{b+a(q+1)-b}{b(q+1)}$
= $\frac{a(q+1)}{b(q+1)}$
= $\frac{a}{b}$
c, $\frac{1}{3}$ = $\frac{1}{(3+1)}$ + $\frac{1}{3(3+1)}$ = $\frac{1}{4}$ + $\frac{1}{12}$
= $\frac{1}{4}$ + $\frac{1}{12+1}$ + $\frac{1}{12(12+1)}$
= $\frac{1}{4}$ + $\frac{1}{13}$ + $\frac{1}{156}$
Giải thích các bước giải: