\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

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

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

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

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)

\(10\dfrac{1}{5}-5\dfrac{1}{2}\times\dfrac{60}{11}+3\div15\%\\ =\dfrac{51}{5}-\dfrac{11}{2}\times\dfrac{60}{11}+3\div\dfrac{15}{100}\\ =\dfrac{51}{5}-30+20\\ =10,2-30+20\\ =0,2\)

\(\dfrac{5}{7}\times\dfrac{5}{11}+\dfrac{5}{7}\times\dfrac{2}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times-\dfrac{7}{11}=-\dfrac{5}{11}\)

\(-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times1+\dfrac{12}{7}\\ =1\)

b: \(27D=3^{14}+3^{17}+...+3^{2024}\)

\(\Leftrightarrow26D=3^{2024}-3^{11}\)

hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)

c: \(25E=-5^4-5^6-...-5^{1002}\)

\(\Leftrightarrow24E=-5^{1002}+5^2\)

hay \(E=\dfrac{-5^{1002}+5^2}{24}\)

\(\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\times-\dfrac{7}{11}=\dfrac{-5}{11}\)

= 5/7 x ( 5/11 + 2/11 - 14/11 )

= 5/7 x -7/11

= -5/11

5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5/11

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

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 \(B=3\)

8/3 . 2/5 . 3/8 . 10 .19/92 trên 5/7 . 5/11 +5/7 .2/11 - 5/7.11/11

=16/15 .15/4 .19/92 trên 5/7(5/11 +2/11 -11/11)

= .4 .19/92 trên 5/7.(-4/11)

=19/2 trên -20/77

\(\frac{\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}}{\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{11}{11}}\)

=\(\frac{\frac{2}{5}.10.\frac{19}{92}}{\frac{5}{7}.\left(\frac{5}{11}+\frac{2}{11}-1\right)}\)

=\(\frac{\frac{19}{23}}{\frac{-20}{77}}\)

=\(\frac{19}{23}.\frac{-77}{20}\)

=\(\frac{-1463}{460}\)

5/7×5/11+5/7×2/11-5/7×14/11

= 5/7 x ( 5/11 + 2/11 - 14/11 )

= 5/7 x -7/11

= -35/77

a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)

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

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

a ,b

cách 1 : cộng 2 số trong ngặc rồi tính bình thường 

cách 2 : nhân phân phối ra rồi tính nhân chia trước cộng trừ sau

c,d

cách 1 : tính bình thường, nhân chia trc cộng trừ sau

cách 2 : đặt nhân tử ( thừa số ) chung

Cách 1 : \((\frac{6}{11}+\frac{5}{11})\cdot\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}\)

Cách 2 : \(\frac{6}{11}\cdot\frac{3}{7}+\frac{5}{11}\cdot\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)

Cách 1 dễ rồi nha

Cách 2 : \(\frac{6}{7}:\frac{2}{5}-\frac{4}{7}:\frac{2}{5}=\frac{15}{7}-\frac{10}{7}=\frac{5}{7}\)

Cách 1 tự làm

Cách 2 : \(\frac{3}{5}\cdot(\frac{7}{9}-\frac{2}{9})=\frac{3}{5}\cdot\frac{5}{9}=\frac{1}{3}\)

Cách 1 : ...

Cách 2 : \((\frac{8}{15}+\frac{7}{15}):\frac{2}{11}=1:\frac{2}{11}=\frac{2}{11\cdot1}=\frac{2}{11}\)