\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)

\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)

a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)

b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)

\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

c) -55/36

d)-67/120 

1, \(\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}=\left(\dfrac{2}{3}-\dfrac{3}{2}\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{6}\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{8}+\dfrac{1}{2}=-\dfrac{1}{8}\)

2, \(\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=\left(-1\right)+1-\dfrac{3}{7}=0-\dfrac{3}{7}=-\dfrac{3}{7}\)

\(\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}=\dfrac{-5}{6}x\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{-5}{8}+\dfrac{1}{2}=\dfrac{-1}{8}\)

\(\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=-1+1-\dfrac{3}{7}=0-\dfrac{3}{7}=\dfrac{-3}{7}\)

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

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

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

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

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

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

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

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

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

Giải:

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

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

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

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

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

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

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

`(6-2 4/5)*3 1/8-1 3/5:1/4`

`=(6-14/5)*25/8-8/5*4`

`=16/5*25/8-32/5`

`=10-32/5=18/5`

=(30/5-14/5).25/8.4

=16/5.25/2

=8.5/1.1

=40

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}=\dfrac{5}{12}.\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)=\dfrac{5}{12}.\left(-1\right)=-\dfrac{5}{12}\)

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}.\left(-\dfrac{9}{2}\right)=-\dfrac{3}{2}\)

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

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

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

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

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