a, \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\\ =-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\\ =\dfrac{-5.4+3.3-1.6+13.2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

b, \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}.82\dfrac{1}{2}\\ =\dfrac{11}{50}.\left(-17\dfrac{1}{2}-82\dfrac{1}{2}\right)=\dfrac{11}{50}.\left(-100\right)=-22\)

a) \(-1\dfrac{2}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(2\dfrac{1}{6}\)

=\(-\dfrac{5}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(\dfrac{13}{6}\)\()\)

=\(-\) \(\dfrac{20}{12}\) + \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) + \(\dfrac{26}{12}\)

= \((\)\(\dfrac{-20}{12}\) + \(\dfrac{26}{12}\) \()\) + \((\) \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) \()\)

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

= \(\dfrac{3}{4}\)

b)\(\dfrac{11}{50}\) \((\) \(-17\dfrac{1}{2}\) \()\) \(-\) \(\dfrac{11}{50}\) .\(82\dfrac{1}{2}\)

= \(\dfrac{11}{50}\) . \(-\dfrac{35}{2}\) \(-\) \(\dfrac{11}{50}\) . \(\dfrac{165}{2}\)

= \(\dfrac{11}{50}\). \((\) \(-\dfrac{35}{2}\) \(-\) \(\dfrac{165}{2}\) \()\)

=\(\dfrac{11}{50}\). \(-\)\(100\)

= \(-22\)

Chúc bạn học thật tốt nha !

a) \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\)

\(=\dfrac{-5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\)

\(=\dfrac{-20}{12}+\dfrac{9}{12}-\dfrac{6}{12}+\dfrac{26}{12}\)

\(=\dfrac{-20+9-6+26}{12}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}\cdot82\dfrac{1}{2}\)

\(=\dfrac{11}{50}\cdot\dfrac{-35}{2}-\dfrac{11}{50}\cdot\dfrac{165}{2}\)

\(=\dfrac{11}{50}\left(\dfrac{-35}{2}-\dfrac{165}{2}\right)\)

\(=\dfrac{11}{50}\cdot\left(-100\right)\)

\(=-22\)

a) Ta có: \(\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-\dfrac{19}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\dfrac{-9}{16}\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}\cdot\dfrac{-16}{9}\)

\(=\dfrac{-112}{144}=\dfrac{-7}{9}\)

b) Ta có: \(17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)

\(=17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}\cdot\dfrac{4}{27}+350\%\)

\(=\dfrac{4}{27}\left(17+\dfrac{6}{11}-8-\dfrac{6}{11}\right)+\dfrac{7}{2}\)

\(=\dfrac{4}{27}\cdot9+\dfrac{7}{2}\)

\(=\dfrac{4}{3}+\dfrac{7}{2}=\dfrac{8}{6}+\dfrac{21}{6}=\dfrac{29}{6}\)

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: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

1) âm năm phần 12

2) âm mười bảy phần 9

3) -1 

Đây là đáp án còn làm bài từ làm nhé

\(\left(-\dfrac{5}{11}+\dfrac{7}{22}--\dfrac{4}{33}-\dfrac{5}{44}\right):\left(38\dfrac{1}{22}-39\dfrac{7}{22}\right)\)
\(=\left(-\dfrac{60}{132}+\dfrac{42}{132}+\dfrac{16}{132}-\dfrac{15}{132}\right):\left(\dfrac{837}{22}-\dfrac{865}{22}\right)\)
\(=-\dfrac{17}{132}\cdot\left(-\dfrac{14}{11}\right)\)
\(=\dfrac{119}{726}\)
#AvoidMe

\(=\left(\dfrac{-60}{132}+\dfrac{42}{132}+\dfrac{16}{132}-\dfrac{15}{432}\right):\left(-1-\dfrac{3}{11}\right)\)

\(=\dfrac{-17}{132}\cdot\dfrac{-11}{14}=\dfrac{17}{168}\)

=-17/132*-11/14=17/168

a)

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

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =3-2\\ =1\)

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)

a: \(=\left(\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

=(căn 3-căn 2)(căn 3+căn 2)

=3-2=1

b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)

=(2-căn 11)(2+căn 11)

=4-11

=-7

a, = (58/9 + 7/11) - (40/9 - 26/11)

= 701/99 - 206/99

= 5

b, = 51/5 - 11/2 . 60/11 + 3 : 3/20

= 51/5 - 30 + 20

= -99/5 + 20

= 1/5

c, = 19/4 + (-0,37) + 1/8 + (-1,8) + (-2,5) + 37/12

= 219/50 + -67/40 + 7/12

= 1973/600

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

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

a: =11/7(-3/7+4/11-4/7+7/11)=0

b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)

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

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

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

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

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

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

                             \(x=10\)