\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

5, \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)

= \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)

= \(\dfrac{1}{7}\).1

=\(\dfrac{1}{7}\)

5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\) = \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)

= \(\dfrac{1}{7}.\dfrac{7}{5}=\dfrac{1}{5}\)

\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)

1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.

2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.

3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).

e: \(\frac25\cdot\frac13-\frac{2}{15}:\frac15+\frac35\cdot\frac13\)

\(=\frac{2}{15}+\frac{3}{15}-\frac{2}{15}\cdot5\)

\(=\frac{5}{15}-\frac{10}{15}=-\frac{5}{15}=-\frac13\)

d: \(19\frac58:\frac{7}{12}-15\frac14:\frac{7}{12}\)

\(=\left(19+\frac58-15-\frac14\right):\frac{7}{12}\)

\(=\left(4+\frac58-\frac28\right):\frac{7}{12}=\left(4+\frac38\right)\cdot\frac{12}{7}=\frac{35}{8}\cdot\frac{12}{7}=5\cdot\frac32=\frac{15}{2}\)

a: \(2\frac37+\left(\frac29-1\frac37\right)-\frac53:\frac19\)
\(=2+\frac37+\frac29-1-\frac37-\frac53\cdot9\)

\(=1+\frac29-5\cdot3=1-15+\frac29=-14+\frac29=\frac{-126+2}{9}=-\frac{124}{9}\)

Nhân nghịch đảo - phân phối- rút gọn

P=(2/3 -1/4+5/11):(5/12+1-7/11)

P=(88/132-33/132+60/132):(55/132+132/132-84/132)

P=115/132:103/132

P=115/103

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)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)

a: =-1/3+1/3=0

b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)

c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)

d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)

a: =-1/3+1/3=0

b: 

c: 

d: 

a: \(\frac{-1}{9}\cdot\frac{-3}{5}+\frac{5}{-6}\cdot\frac{-3}{5}-\frac72\cdot\frac35\)

\(=\frac19\cdot\frac35+\frac56\cdot\frac35-\frac72\cdot\frac35\)

\(=\frac35\left(\frac19+\frac56-\frac72\right)\)

\(=\frac35\left(\frac{2}{18}+\frac{15}{18}-\frac{63}{18}\right)=\frac35\cdot\frac{-46}{18}=\frac35\cdot\frac{-23}{9}=\frac{-23}{3\cdot5}=-\frac{23}{15}\)

b: Sửa đề: \(-\frac37\cdot\frac{15}{13}-\frac37\cdot\frac{11}{13}-\frac37\)

\(=-\frac37\left(\frac{15}{13}+\frac{11}{13}\right)-\frac37\)

\(=-\frac37\cdot2-\frac37=-\frac37\cdot3=-\frac97\)