A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)⁰

= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1

= \(\dfrac{11}{9}\)(-9) + 1

= -11 + 1

= -10

= \(\dfrac{11}{15}\). \(\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\)

= \(\dfrac{11}{15}\). 1

=\(\dfrac{11}{15}\)

\(=\dfrac{11}{15}\cdot\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)=\dfrac{11}{15}\cdot\dfrac{11}{11}=\dfrac{11}{15}\)

11/15 . ( 4/11 +5/11 + 2/11 )

= 11/15 . 1

= 11/15

`4 / 15 . 1 / 3 . 15 / 20`

`= [ 4 . 1 . 15 ] / [ 15 . 3 . 20 ]`

`= [ 4 . 1 . 15 ] / [ 15 . 3 . 4 . 5 ]`

`= 1 / [ 3 . 5 ]`

`= 1 / 15`

\(=\left(\dfrac{4}{15}.\dfrac{15}{20}\right).\dfrac{1}{3}=\dfrac{60}{300}.\dfrac{1}{3}=\dfrac{60}{900}=\dfrac{1}{15}\)

\(=\dfrac{4.1.15}{15.3.5.4}=\dfrac{1}{3.5}=\dfrac{1}{15}\)

\(=\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=-\dfrac{6}{5}\times\dfrac{5}{7}+\dfrac{13}{5}=\dfrac{61}{35}\)

=\(\dfrac{-3}{5}\).\(\dfrac{5}{7}\)-\(\dfrac{3}{5}\).\(\dfrac{5}{7}\)+\(\dfrac{13}{5}\)

=(\(\dfrac{-3}{5}\)-\(\dfrac{3}{5}\)+\(\dfrac{13}{5}\)).\(\dfrac{5}{7}\)
=\(\dfrac{13}{5}\).\(\dfrac{5}{7}\)
=\(\dfrac{13}{1}\).\(\dfrac{1}{7}\)
=\(\dfrac{13}{7}\)

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)

=\(\left(4-\dfrac{5}{12}\right).\dfrac{1}{2}+\dfrac{5}{24}\)

=\(4.\dfrac{1}{2}-\dfrac{5}{12}.\dfrac{1}{2}+\dfrac{5}{24}\)

=\(2-\dfrac{5}{24}+\dfrac{5}{24}=2\)

Giải:

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\) 

\(=\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\) 

\(=\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\) 

\(=\dfrac{1}{3}.\left(\dfrac{2}{5}-2+\dfrac{3}{5}\right)\) 

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\) 

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)

\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)

\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)

\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)

\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)

a. 3/5+1/5.-17/9

=3/5+-17/45

=27/45+-17/45

=2/9

b. (-4/15-18/19)-(20/19+11/15)

=-4/15-18/19-20/19-11/15

=(-4/15-11/15)-(18/19-20/19)

=(-1)-(-2/19)

=(-19/19)-(-2/19)

=-17/19  

a.2/9
b.-3

a.2/9 b.(-3)

a.2/9
b.-3

a)\(=\dfrac{3}{6}-\dfrac{4}{6}=-\dfrac{1}{6}\)

b)\(=-\dfrac{35}{9}-\dfrac{10}{9}=-\dfrac{45}{9}=-5\)

c)\(=\left[-10+9\right].\left[1,3.\left(2,5+7,5\right)\right]=-1.\left[1,3.10\right]=-1.13=-13\)