Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.-1\\ b.\ 1\\ c.\ 1\\ d.\ -\frac{4}{3}\\ Bài\ 2:\\ a.\ S=\{-15\}\\ b.\ \ S=\left\{\frac{3}{2}\right\}\\ c.S=\left\{-\frac{7}{2}\right\}\\ d.S=\{17\}\\ e.\ \ S=\left\{\frac{7}{6}\right\}\\ f.\ \ S=\left\{-\frac{1}{2}\right\} \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.=\frac{16}{3}\left(\frac{10}{24} -\frac{9}{24}\right) -\frac{11}{9} =\frac{16}{3} .\frac{1}{24} -\frac{11}{9} =\frac{2}{9} -\frac{11}{9} =-1\\ b.\ =\frac{5}{17}\left(\frac{40}{15} -\frac{6}{15}\right) +\frac{2}{7}\left(\frac{9}{6} -\frac{2}{6}\right) =\ \frac{5}{17} .\frac{34}{15} +\frac{2}{7} .\frac{7}{6}\\ =\frac{2}{3} +\frac{1}{3} =1\\ c.\ -\frac{3}{40} :\left(\frac{6}{20} -\frac{16}{20} +\frac{3}{20}\right) +\frac{11}{14} =-\frac{3}{40} :\frac{-7}{20} +\frac{11}{14} =\frac{3}{14} +\frac{11}{14} =1\\ d.\ \frac{4}{3} -\frac{10}{9}\left[\frac{18}{5}\left(\frac{-1}{6}\right)^{2} +\frac{23}{10}\right] =\frac{4}{3} -\frac{10}{9}\left[\frac{1}{10} +\frac{23}{10}\right] =\frac{4}{3} -\frac{10}{9} .\frac{24}{10}\\ =\frac{4}{3} -\frac{8}{3} =-\frac{4}{3}\\ Bài\ 2:\\ a.\ \Leftrightarrow 4x=-60\Leftrightarrow x=-15\\ Vậy\ S=\{-15\}\\ b.\ \Leftrightarrow -\frac{6}{10} x+\frac{3}{10} x=-\frac{9}{20} \Leftrightarrow -\frac{3}{10} x=-\frac{9}{20} \Leftrightarrow x=\frac{3}{2}\\ Vậy\ S=\left\{\frac{3}{2}\right\}\\ c.\ \frac{5}{12} x+\frac{8}{12} x-\frac{15}{12} x=\frac{7}{12} \Leftrightarrow -2x=7\Leftrightarrow x=-\frac{7}{2}\\ Vậy\ S=\left\{-\frac{7}{2}\right\}\\ d.\ \frac{28}{40} +\frac{14}{40} x-\frac{15}{40} x=\frac{3}{40} \Leftrightarrow -\frac{1}{40} x=-\frac{17}{40} \Leftrightarrow x=17\\ Vậy\ S=\{17\}\\ e.\ \frac{14}{5}\left(\frac{21}{20} x-\frac{11}{10}\right) =\frac{7}{20} \Leftrightarrow \frac{21}{20} x-\frac{11}{10} =\frac{1}{8}\\ \Leftrightarrow \frac{21}{20} x=\frac{49}{40} \Leftrightarrow 42x=49\Leftrightarrow x=\frac{7}{6}\\ Vậy\ S=\left\{\frac{7}{6}\right\}\\ f.\ \frac{10}{9}\left(\frac{31}{12} x+\frac{5}{3}\right) =\frac{5}{12} \Leftrightarrow \frac{31}{12} x+\frac{5}{3} =\frac{3}{8}\\ \Leftrightarrow \frac{31}{12} x=-\frac{31}{24} \Leftrightarrow x=-\frac{1}{2}\\ Vậy\ S=\left\{-\frac{1}{2}\right\} \end{array}$

Đáp án:

$\\$

Bài `1.`

`d,`

`4/3 - 10/9 [18/5 (4/3 - 3/2)^2 + 23/10]`

`= 4/3 - 10/9 [18/5 (-1/6)^2 + 23/10]`

`= 4/3 - 10/9 [18/5 . 1/36 + 23/10]`

`= 4/3 - 10/9 [1/10 + 23/10]`

`= 4/3 - 10/9 . 24/10`

`= 4/3 - 8/3`

`= (-4)/3` 

Bài `2.`

`e,`

`14/5 (x:20/21 - 11/10) + 2/5 = 3/4`

`-> 14/5 (x :20/21 - 11/10) = 3/4 - 2/5`

`-> 14/5 (x : 20/21 - 11/10) = 7/20`

`-> x : 20/21 - 11/10 = 7/20 : 14/5`

`-> x : 20/21 - 11/10=1/8`

`-> x : 20/21 = 1/8 + 11/10`

`-> x : 20/21=49/40`

`->x=49/40 . 20/21`

`->x=7/6`

Vậy `x=7/6`

`f,`

`10/9 (31/12x + 5/3) - 7/12=(-1)/6`

`->10/9 (31/12x + 5/3) = (-1)/6 + 7/12`

`-> 10/9 (31/12x + 5/3)=5/12`

`-> 31/12x + 5/3 = 5/12 : 10/9`

`-> 31/12x + 5/3=3/8`

`-> 31/12x = 3/8 - 5/3`

`-> 31/12x = (-31)/24`

`->x=(-31)/24 : 31/12`

`->x=(-1)/2`

Vậy `x=(-1)/2`