\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

Giải:

1) (-8/13:3/7+-5/13:3/7).(-4)³.|-3|/7

=[7/3.(-8/13+-5/13)].-192/7

=[7/3.(-1)].-192/7

=-7/3.-192/7

=64

2) 75%-(5/2+5/3)+(-1/2)²

=3/4-25/6+1/4

=(3/4+1/4)-25/6

=1-25/6

=-19/6

Chúc bạn học tốt!

1) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right).|-3|}{7}\)

  = \(\left[\left(\dfrac{-8}{13}+\dfrac{-5}{13}\right):\dfrac{3}{7}\right].\dfrac{-64.3}{7}\)

  = \(\left[-1:\dfrac{3}{7}\right].\dfrac{-192}{7}\)

  = \(\dfrac{-7}{3}.\dfrac{-192}{7}\)

  =       \(64\)

2)  \(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)

  = \(\dfrac{3}{4}-\dfrac{25}{6}+\dfrac{1}{4}\)

  = \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)-\dfrac{25}{6}\)

  =          \(1-\dfrac{25}{6}\)

  =            \(\dfrac{-19}{6}\)

Chúc bạn học tốt !

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

\(=\left(\dfrac{-3}{5}-\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)

\(=\dfrac{-6}{5}:\dfrac{7}{5}+\dfrac{13}{5}\)

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

2, \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\left(-\dfrac{1}{9}\right)\)

\(=0,75-\dfrac{37}{12}+9\cdot\left(-\dfrac{1}{9}\right)\)

\(=-\dfrac{7}{3}+\left(-1\right)\\ -\dfrac{10}{3}\)

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

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

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

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

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

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

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

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

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

Giải:

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

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

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

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

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

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

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

Lời giải :

a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=2,5\)

b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

\(=\dfrac{3}{7}\left(19-33\right)\)

\(=\dfrac{3}{7}\left(-14\right)\)

\(=-6\)

c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)

\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)

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

\(=0\)

d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)

\(=-10\left(-\dfrac{7}{5}\right)\)

\(=14\)

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

Câu 50 tus sửa đề 

\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)

Bài 1 :

Câu 49 . = 38/5 - ( 19/7 + 28/5 )

               = 38/5 - 19/7 - 28/5 

               = ( 38/5 - 28/5 ) - 19/7 

               = 2 - 19/7 = -5/7

Câu 50 . = ( -1/9 : -25/9 ) . 15/22

               = 1/25 . 15/22 = 3/110 

Đúng 100 % nha bạn 🤍

mình no bik!

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

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

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

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

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