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) 

\(\dfrac{3}{4}+\dfrac{-7}{5}+\dfrac{1}{4}+\dfrac{-3}{5}\)

\(=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{-7}{5}-\dfrac{3}{5}\right)\)

\(=1-2\)

\(=-1\)

b)

\(\dfrac{4}{9}.\dfrac{7}{11}-\dfrac{4}{11}.\dfrac{2}{9}+\dfrac{6}{11}.\dfrac{4}{9}\)

\(=\dfrac{28}{99}-\dfrac{8}{99}+\dfrac{24}{99}\)

\(=\dfrac{28-8+24}{99}\)

\(=\dfrac{44}{99}\)

\(=\dfrac{4}{9}\)

a) \(\dfrac{7}{11}+\dfrac{3}{4}+\dfrac{4}{11}+\dfrac{1}{4}=\left(\dfrac{7}{11}+\dfrac{4}{11}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=1+1=2\)

b) \(\dfrac{72}{99}-\dfrac{28}{99}-\dfrac{14}{99}=\dfrac{72-28-14}{99}=\dfrac{30}{99}=\dfrac{10}{33}\)

c) \(69.78+35.97+30.22=\left(69.78+30.22\right)+35.97=100+35.97=135.97\)

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: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

\(\dfrac{4}{11}< \dfrac{6}{11}\)

\(\dfrac{6}{7}=\dfrac{12}{14}\)

\(\dfrac{15}{17}>\dfrac{10}{17}\)

\(\dfrac{2}{3}< \dfrac{3}{4}\)

67=121467=1214

23<34

4/11 < 6/11

15/17 > 10/17

6/7 = 12/14 

2/3< 3/4

a ( ....< ....)

b( ....<....)

a <

b <

tick cho mk nha