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a, Ta có: \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{n.\left(n+4\right)}=\dfrac{1}{4}.\dfrac{4}{n.\left(n+1\right)}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
Vậy \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
b, \(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}=4.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4.\dfrac{32}{99}=\dfrac{128}{99}\)
Vậy \(A=\dfrac{128}{99}\)
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
a) 3/4 + (-7/5) + 1/4 + (-3/5)
= (3/4 + 1/4) + (-7/5 - 3/5)
= 1 - 2
= -1
b) 4/9 . 7/11 - 4/11 . 2/9 + 6/11 . 4/9
= 4/9 . (7/11 - 2/11 + 6/11)
= 4/9 . 1
= 4/9
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
\(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\\ \dfrac{-15}{48}-\dfrac{1}{12}=\dfrac{-5}{16}-\dfrac{1}{12}=\dfrac{-15}{48}-\dfrac{4}{48}=\dfrac{-19}{48}\\ \dfrac{-3}{4}-\dfrac{4}{5}=\dfrac{-15}{20}-\dfrac{16}{20}=\dfrac{-31}{20}\\ \dfrac{3}{4}+\dfrac{-5}{6}-\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}-\dfrac{11}{12}=\dfrac{-12}{12}=-1\)
128/99
A = \(\dfrac{4}{3}\) . \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) . \(\dfrac{4}{11}\) + \(\dfrac{4}{11}\) . \(\dfrac{4}{15}\) + ... + \(\dfrac{4}{95}\) . \(\dfrac{4}{99}\)
A = \(\dfrac{4.4}{3.7}\) + \(\dfrac{4.4}{7.11}\) + \(\dfrac{4.4}{11.15}\) + ... + \(\dfrac{4.4}{95.99}\)
A = \(\dfrac{16}{3.7}\) + \(\dfrac{16}{7.11}\) + \(\dfrac{16}{11.15}\) + ... + \(\dfrac{16}{95.99}\)
A = 4.( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{95.99}\))
A = 4.( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{1}{3}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{33}{99}\) + \(\dfrac{-1}{99}\))
A = 4. \(\dfrac{32}{99}\)
A = \(\dfrac{4.32}{99}\)
A = \(\dfrac{128}{99}\)
Vậy A = \(\dfrac{128}{99}\)