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

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: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)

9: =2/7-2/5+5/7=1-2/5=3/5

10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)

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

8 = -2/11

9 = 3/5

10 = -5/19

11 = -5/7

11 = 5/13

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

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

Mik làm Bài 2 nhé ~

Bài 2 :

a) \(x-\dfrac{1}{2}=-\dfrac{1}{10}\)

\(x=-\dfrac{1}{10}+\dfrac{1}{2}\)

\(x=\dfrac{2}{5}\)

b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)

\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)

\(\dfrac{2}{3}x=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}:\dfrac{2}{3}\)

\(x=\dfrac{11}{3}.\dfrac{3}{2}\)

\(x=\dfrac{11}{2}\)

c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)

\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=2,5-\dfrac{3}{4}\)

\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{5}{2}-\dfrac{3}{4}\)

\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)

\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\dfrac{1}{8}x=\dfrac{5}{4}\)

\(x=10\)

Bài 1:

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

\(=\dfrac{-4}{11}.\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{7}{11}\) 

\(=\dfrac{-4}{11}.1-\dfrac{7}{11}\) 

\(=\dfrac{-4}{11}-\dfrac{7}{11}\) 

\(=-1\) 

b) \(\dfrac{3}{5}:\dfrac{-7}{10}+0,5-\left(\dfrac{-9}{14}\right)\) 

\(=\dfrac{-6}{7}+\dfrac{1}{2}+\dfrac{9}{14}\) 

\(=\dfrac{2}{7}\) 

c) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\) 

\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\) 

\(=\dfrac{3}{5}-\dfrac{8}{5}:6\) 

\(=\dfrac{3}{5}-\dfrac{4}{15}\) 

\(=\dfrac{1}{3}\)

Bài 2:

a) \(x-\dfrac{1}{2}=\dfrac{-1}{10}\) 

            \(x=\dfrac{-1}{10}+\dfrac{1}{2}\) 

            \(x=\dfrac{2}{5}\) 

b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\) 

            \(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\) 

            \(\dfrac{2}{3}x=\dfrac{11}{3}\) 

               \(x=\dfrac{11}{3}:\dfrac{2}{3}\) 

               \(x=\dfrac{11}{2}\) 

c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\) 

                   \(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{5}{2}-\dfrac{3}{4}\) 

                   \(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\) 

                          \(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\) 

                          \(\dfrac{1}{8}x=\dfrac{5}{4}\) 

                             \(x=\dfrac{5}{4}:\dfrac{1}{8}\) 

                             \(x=10\)

cậu lấy máy tính,tính đi ạ

-1,11765149

Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)

a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)

d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)

a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)

\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)

\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

a,\(=\dfrac{2}{9}+\left(\dfrac{-3}{10}+\dfrac{-7}{10}\right)=\dfrac{2}{9}+\left(-1\right)=\dfrac{-7}{9}\)

b,\(=\left(\dfrac{-11}{6}+\dfrac{-1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-8}{5}\)

c,\(=\dfrac{27}{13}-\left(\dfrac{106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)

a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)

\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)

\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)

a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)

b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)

d) \(\left(\dfrac{3}{4}+0,5+25\%\right).2\dfrac{2}{3}=\left(\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{4}\right).\dfrac{8}{3}=\left(\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}\right).\dfrac{8}{3}=\dfrac{3}{2}.\dfrac{8}{3}=4\)