Đáp án:

 

Giải thích các bước giải:

a/ $D=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{19.20}$

⇔ $D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{19}-\dfrac{1}{20}$

⇔ $D=\dfrac{1}{2}-\dfrac{1}{20}$

⇔ $D=\dfrac{10}{20}-\dfrac{1}{20}$

⇔ $D=\dfrac{9}{20}$

b/ $E=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}$

⇔ $E=\dfrac{1}{99}-(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{97.98}+\dfrac{1}{98.99})$

⇔ $E=\dfrac{1}{99}-(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99})$

⇔ $E=\dfrac{1}{99}-(1-\dfrac{1}{99})$

⇔ $E=\dfrac{1}{99}-1+\dfrac{1}{99}$

⇔ $E=\dfrac{2}{99}-1$

⇔ $E=-\dfrac{97}{99}$

$\text{Quy tắc cho bài trên:}$

$\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$

$\text{Quy tắc chung:}$

$\dfrac{1}{k.(k+a)}=\dfrac{1}{a}.(\dfrac{1}{k}-\dfrac{1}{k+a})$

Chúc bạn học tốt !!!

Giải thích các bước giải:

 `D=1/2.3+1/3.4+...+1/19.20`

`=1/2-1/3+1/4-1/4+...+1/19-1/20`

`=1/2-1/20`

`=9/20`

 `E=1/99-1/99.98-1/98.97-...-1/2.1`

`=1/99-(1/1.2+1/2.3+...+1/98.99)`

`=1/99-(1-1/2+1/2-1/3+...+1/98-1/99)`

`=1/99-(1-1/99)`

`=1/99-98/99`

`=-97/99`