Задачи от събиране и изваждане на дроби - много трудни задачи с решения

Решение:
[tex]\frac{3}{4} - \frac{1}{2}(\frac{1}{3} + \frac{3}{7}) = \frac{3}{4} - \frac{1}{2}(\frac{1 \times 7}{3 \times 7} + \frac{3 \times 3}{7\times3}) = [/tex] [tex]\frac{3}{4} - \frac{1}{2}(\frac{7 + 9}{21})= \frac{3}{4} - \frac{1}{2}.\frac{16}{21} = \frac{3}{4} - \frac{8}{21} = [/tex] [tex]\frac{3\times21}{4\times21} - \frac{8\times4}{21\times4} = \frac{63 - 32}{84} = \frac{31}{48}[/tex]

Решение:
[tex]\frac{1}{2}+\frac{5}{13}+\frac{1}{26}=\frac{1.13}{2.13}+\frac{5.2}{2.13}+\frac{1}{26}=\frac{13}{26}+\frac{10}{26}+\frac{1}{26}=\frac{13+10+1}{26}=\frac{24}{26}=\frac{12}{13}[/tex]

Решение:
[tex]\frac{5}{6}-\frac{7}{12}+\frac{1}{24}=\frac{4.5}{4.6}-\frac{7.2}{12.2}+\frac{1}{24}=\frac{20}{24}-\frac{14}{24}+\frac{1}{24}=\frac{20-14+1}{24}=\frac{7}{24}[/tex]

Решение:
[tex]\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}=\frac{3.5}{4.5}-\frac{4.4}{5.4}+\frac{5.7}{6.7}-\frac{6.6}{6.7}=(\frac{15}{20}-\frac{16}{20})+(\frac{35}{42}-\frac{36}{42})=\frac{-1}{20}+\frac{-1}{42}=\frac{-1.21}{20.21}+\frac{-1.10}{42.10}=\frac{-21}{420}+\frac{-10}{420}=\frac{-21-10}{420}=-\frac{31}{420}[/tex]

Решение:
[tex]\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81} = \frac{1}{3}.1+\frac{1}{3}.\frac{1}{3}+\frac{1}{27}.1+\frac{1}{27}.\frac{1}{3}=\frac{1}{3}(1+\frac{1}{3})+\frac{1}{27}(1+\frac{1}{3})=(\frac{1}{3}+\frac{1}{27})(\frac{1}{3}+1) = (\frac{9}{27}+\frac{1}{27})(\frac{3}{3}+\frac{1}{3})=\frac{10}{27}.\frac{4}{3}=\frac{40}{81}[/tex]

Решение:
[tex]\frac{1}{15}+\frac{3}{20}-\frac{17}{25}+2=\frac{20}{20.15}+\frac{3.15}{20.15}+\frac{2.25}{25}-\frac{17}{25}=\frac{20}{300}+\frac{45}{300}+\frac{50}{25}-\frac{17}{25}=\frac{65}{300}+\frac{33}{25}=\frac{65}{300}+\frac{12.33}{12.25}=\frac{65}{300}+\frac{396}{300}=\frac{65+396}{300}=\frac{461}{300}[/tex]

Решение:
Нека означим [tex]S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...[/tex]. Ако умножим равенството с 2, получаваме [tex]2S=2.1+2.\frac{1}{2}+2.\frac{1}{4}+2.\frac{1}{8}+...=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=2+(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...)=2+S[/tex]. Вземайки първото и последното, виждаме, че [tex]2S=2+S[/tex], или [tex]S=2[/tex]

Решение:
Нека разгледаме израза [tex]\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}[/tex], откъдето виждаме, че [tex]\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}[/tex]. Прилагаме го за [tex]n=2,3,4,5[/tex] и заместваме в израза от условието:
[tex]\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\frac{1}{2}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}[/tex]

Решение:
[tex]\frac{1}{5}+\frac{2}{9}+\frac{7}{15}=\frac{3}{3.5}+\frac{7}{15}+\frac{2}{9}=\frac{3+7}{15}+\frac{2}{9}=\\=\frac{10}{15}+\frac{2}{9}=\frac{2}{3}+\frac{2}{9}=\frac{2.3}{3.3}+\frac{2}{9}=\frac{6}{9}+\frac{2}{9}=\frac{8}{9}[/tex]

Решение:
[tex]\frac{3}{27}+\frac{2.11}{2.15}+\frac{17}{30}=\frac{3}{3.9}+\frac{22+17}{30}=\frac{1}{9}+\frac{39}{30}=\frac{1}{9}+\frac{13}{10}=\frac{10}{90}+\frac{9.13}{90}=\frac{10+117}{90}=\frac{127}{90}[/tex]

Решение:
[tex]\frac{5}{14}+\frac{3}{28}+\frac{13}{21}=\frac{10}{28}+\frac{3}{28}+\frac{13}{21}=\frac{13}{28}+\frac{13}{21}=\frac{13}{7}.\frac{1}{4}+\frac{13}{7}.\frac{1}{3}=\\=\frac{13}{7}(\frac{1}{4}+\frac{1}{3})=\frac{13}{7}(\frac{3}{12}+\frac{4}{12})=\frac{13}{7}.\frac{7}{12}=\frac{13}{12}[/tex]

Решение:
[tex]\frac{13}{19}+\frac{5}{7}+\frac{8}{11}=\frac{13.7.11}{19.7.11}+\frac{5.19.11}{7.19.11}+\frac{8.19.7}{7.19.11}=\\=\frac{13.7.11+5.19.11+8.7.19}{7.11.19}=\frac{13.77+19(5.11+8.7)}{7.11.19}=\frac{1001+19(56+55)}{77.19}=\frac{1001+19.111}{1463}=\frac{3110}{1463}[/tex]

Решение:
[tex]\frac{5}{14}-\frac{8}{21}+\frac{17}{35}=\frac{3.5}{3.14}-\frac{2.8}{2.21}+\frac{17}{35}=\frac{15-16}{42}+\frac{17}{35}=\frac{17}{35}-\frac{1}{42}=\frac{6.17}{35.6}-\frac{5}{42.5}=\frac{102}{210}-\frac{5}{210}=\frac{97}{210}[/tex]

Решение:
[tex]\frac{7}{15}-\frac{8}{35}+\frac{9}{21}=\frac{7.7}{15.7}-\frac{8.3}{35.3}+\frac{9}{21}=\frac{49-24}{105}+\frac{9}{21}=\frac{25}{105}+\frac{9}{21}=\frac{5}{21}+\frac{9}{21}=\frac{14}{21}=\frac{2}{3}[/tex]

Решение:
[tex]\frac{5}{49}+\frac{3}{3.7}-\frac{6}{25}=\frac{5}{49}+\frac{1}{7}-\frac{6}{25}=\frac{5}{49}+\frac{7}{49}-\frac{12}{50}=[/tex]
[tex]=\frac{12}{49}-\frac{12}{50}=12(\frac{1}{49}-\frac{1}{50})=\frac{12}{49.50}=\frac{6}{49.25}=\frac{6}{1225}[/tex]