Đáp án:

Giải thích các bước giải:

 Bài 1 :

$a)\dfrac{1}{12}+$ $\dfrac{-3}{12}=$ $\dfrac{1-3}{12}=$ $\dfrac{-2}{12}=$ $\dfrac{-1}{6}$

$b)\dfrac{7}{8}-$ $\dfrac{5}{4}=$ $\dfrac{7}{8}-$$\dfrac{10}{8}=$ $\dfrac{7-10}{8}=$ $\dfrac{-3}{8}$

$c)1\dfrac{2}{5}+$ $3\dfrac{3}{5}=$ $\dfrac{7}{5}+$ $\dfrac{18}{5}=$ $\dfrac{25}{5}=5$

$d)\dfrac{-14}{20}+0,6=$ $\dfrac{-7}{10}+$ $\dfrac{6}{10}=$ $\dfrac{-1}{10}$

$e)\dfrac{7}{3}-($ $\dfrac{-1}{4}-$ $\dfrac{5}{12})=$ $\dfrac{7}{3}-($ $\dfrac{-3}{12}-$ $\dfrac{5}{12})=$$\dfrac{7}{3}-$$\dfrac{-8}{12}=$ $\dfrac{7}{3}+$$\dfrac{2}{3}=$ $\dfrac{9}{3}=3$

Bài 2 :

$a)x+\dfrac{1}{5}=$ $\dfrac{3}{7}$

→$x=$ $\dfrac{3}{7}-\dfrac{1}{5}$

→$x=\dfrac{15-7}{35}=$ $\dfrac{8}{35}$ 

$b)x-\dfrac{3}{4}=$ $\dfrac{1}{2}$

→$x=$ $\dfrac{1}{2}+\dfrac{3}{4}$

→$x=\dfrac{2+3}{4}=$ $\dfrac{5}{4}$ 

$c)\dfrac{1}{20}-(x-\dfrac{8}{5})=\dfrac{1}{10}$

→$x-\dfrac{8}{5}=\dfrac{1}{20}-\dfrac{1}{10}$ 

→$x-\dfrac{8}{5}=\dfrac{-1}{20}$ 

→$x=\dfrac{-1}{20}+\dfrac{8}{5}=\dfrac{31}{20}$ 

$d)\dfrac{11}{12}-(\dfrac{2}{5}+x)=\dfrac{2}{3}$

→$\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}$

→$x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}$ 

Bài 3 :

$A=\dfrac{1}{1.3}+$$\dfrac{1}{3.5}+$$\dfrac{1}{5.7}+...+$ $\dfrac{1}{19.21}$

$=\dfrac{1}{2}($$1-\dfrac{1}{3}+$$\dfrac{1}{3}-$$\dfrac{1}{5}+$ $\dfrac{1}{5}-$$\dfrac{1}{7}+...+$ $\dfrac{1}{19}-$ $\dfrac{1}{21})$

$=\dfrac{1}{2}-(1-\dfrac{1}{21})$

$=\dfrac{1}{2}.\dfrac{20}{21}=$ $\dfrac{10}{21}$

$B=\dfrac{1}{99}-$ $\dfrac{1}{99.98}-$$\dfrac{1}{98.97}-...-$ $\dfrac{1}{3.2}-$$\dfrac{1}{2.1}$

$=\dfrac{1}{99}-$ ($\dfrac{1}{99.98}+$$\dfrac{1}{98.97}+...+$ $\dfrac{1}{3.2}+$$\dfrac{1}{2.1})$

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

$=\dfrac{1}{99}-(\dfrac{1}{99}-1)$

$=1_{}$

$C=\dfrac{4}{1.5}+$$\dfrac{4}{5.9}+...$ $\dfrac{4}{92.96}+$$\dfrac{4}{96.100}$

$=4-\dfrac{4}{5}+$ $\dfrac{4}{5}-$ $\dfrac{4}{9}+....+$$\dfrac{4}{92}-$ $\dfrac{4}{96}+$ $\dfrac{4}{96}-$ $\dfrac{4}{100}$

$=4-\dfrac{4}{100}=$ $\dfrac{99}{25}$

$D=\dfrac{1}{3.4}+$$\dfrac{1}{4.5}+$ $\dfrac{1}{5.6}+....+$$\dfrac{1}{20.21}$

$=\dfrac{1}{3}-\dfrac{1}{4}+$ $\dfrac{1}{4}-$ $\dfrac{1}{5}+$$\dfrac{1}{5}-$ $\dfrac{1}{6}+...+$ $\dfrac{1}{20}-$ $\dfrac{1}{21}$

$=\dfrac{1}{3}-$ $\dfrac{1}{21}=$ $\dfrac{2}{7}$

$E=\dfrac{1}{2.4}+$$\dfrac{1}{4.6}+$$\dfrac{1}{6.8}+...+$ $\dfrac{1}{28.30}$

$=\dfrac{1}{2}($$\dfrac{1}{2}-\dfrac{1}{4}+$$\dfrac{1}{4}-$$\dfrac{1}{6}+$ $\dfrac{1}{6}-$$\dfrac{1}{8}+...+$ $\dfrac{1}{28}-$ $\dfrac{1}{30})$

$=\dfrac{1}{2}-(\dfrac{1}{2}-\dfrac{1}{30})$

$=\dfrac{1}{2}.\dfrac{7}{15}=$ $\dfrac{7}{30}$