a) `6/18+ (-14)/21 = 1/3 - 2/3=-1/3`

b) `3/5 - (-1/2) = 3/5+1/2=11/10`

c) `2/7 : 3/4 = 2/7 . 4/3 = 8/21`

d) `(-28)/33 . (-3)/4 = 28/33 . 3/4=7/11`

a) \(\dfrac{6}{18}+\dfrac{-14}{21}=\dfrac{1}{3}-\dfrac{2}{3}=\dfrac{-1}{3}\)

b) \(\dfrac{3}{5}-\dfrac{-1}{2}=\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{6}{10}+\dfrac{5}{10}=\dfrac{11}{10}\)

c) \(\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)

d) \(\dfrac{-28}{33}\cdot\dfrac{-3}{4}=\dfrac{28}{4}\cdot\dfrac{3}{33}=7\cdot\dfrac{1}{11}=\dfrac{7}{11}\)

a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)

=\(\dfrac{10}{11}.\dfrac{-1}{2}\)

=\(\dfrac{-5}{11}\)

b; 

B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8

B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8

B = \(\dfrac{2}{7}\) - 8

B = \(\dfrac{2}{7}-\dfrac{56}{7}\)

B = - \(\dfrac{54}{7}\)

c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))

C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))

C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\)  - \(\dfrac{588}{105}\)]

C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]

C = - 15.(- \(\dfrac{143}{105}\))

C = \(\dfrac{143}{7}\)

\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)

\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)

\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

Giải:

a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)

\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)

\(\Leftrightarrow x=-\dfrac{79}{210}\)

Vậy \(x=-\dfrac{79}{210}\).

b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)

\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)

\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)

\(\Leftrightarrow3x=\dfrac{67}{75}\)

\(\Leftrightarrow x=\dfrac{67}{75}:3\)

\(\Leftrightarrow x=\dfrac{67}{225}\)

Vậy \(x=\dfrac{67}{225}\).

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Câu 1: D

Câu 2: A

a) $\frac{2}{3} = \frac{{2 \times 7}}{{3 \times 7}} = \frac{{14}}{{21}}$ ;  Giữ nguyên phân số  $\frac{{16}}{{21}}$

$\frac{3}{7} = \frac{{3 \times 3}}{{7 \times 3}} = \frac{9}{{21}}$

Vì $\frac{9}{{21}} < \frac{{14}}{{21}} < \frac{{16}}{{21}}$ nên các phân số đã cho xếp theo thứ tự từ bé đến lớn là: $\frac{3}{7};\,\,\frac{2}{3};\,\,\,\frac{{16}}{{21}}$

b) $\frac{2}{9} = \frac{{2 \times 3}}{{9 \times 3}} = \frac{6}{{27}}$, Giữ nguyên phân số $\frac{4}{{27}}$

$\frac{1}{3} = \frac{{1 \times 9}}{{3 \times 9}} = \frac{9}{{27}}$

Vì $\frac{4}{{27}} < \frac{6}{{27}} < \frac{9}{{27}}$ nên $\frac{4}{{27}}$< $\frac{2}{9} < \frac{1}{3}$

Vậy các phân số đã cho xếp theo thứ tự từ bé đến lớn là $\frac{4}{{27}}$; $\frac{2}{9};\frac{1}{3}$

c) Giữ nguyên phân số $\frac{{11}}{{28}}$

$\frac{3}{4} = \frac{{3 \times 7}}{{4 \times 7}} = \frac{{21}}{{28}}$ ;  $\frac{2}{7} = \frac{{2 \times 4}}{{7 \times 4}} = \frac{8}{{28}}$

Vì $\frac{8}{{28}} < \frac{{11}}{{28}} < \frac{{21}}{{28}}$ nên các phân số đã cho xếp theo thứ tự từ bé đến lớn là $\frac{2}{7};\frac{{11}}{{28}};\frac{3}{4}$

Hoàng Ngọc Anh bài này nè bn giúp mk nha!!! ngày mai mk phải nộp bài rùi =.=

a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)

\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)

\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)

\(\Rightarrow x=\dfrac{-79}{210}\)

b) Tương tự câu a)

rêtrt

A= 5/4

B= 63/52