\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)

=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)  

=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\) 

=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\) 

=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\) 

=\(-\dfrac{4}{7}-\dfrac{3}{10}\) 

=\(-\dfrac{40}{70}-\dfrac{21}{70}\)

=\(-\dfrac{61}{70}\)

   (3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))

= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)

= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) +  \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)

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

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

= -7 + 3 + \(\dfrac{5}{6}\)

= - 4 + \(\dfrac{5}{6}\)

= \(\dfrac{-19}{6}\)

Đề sai rồi bn, sao lại 1/1/3 được

có thể đó là hỗn số thì sao hả princess angel

a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)

b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)

a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)

c: \(=\left(0.125\right)^{650}\cdot8^{102}\)

\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)

\(=\dfrac{1}{8^{548}}\)

A = 1/25

B = 1

C = 4/5

a) \(1\dfrac{7}{20}:2,7+2,7:1,35+\left(0,4:2\dfrac{1}{2}\right).\left(4,2-1\dfrac{3}{40}\right)\)

\(=\dfrac{27}{20}:\dfrac{27}{10}+\dfrac{27}{10}:\dfrac{27}{10}+\left(\dfrac{2}{5}:\dfrac{5}{2}\right).\left(\dfrac{21}{5}-\dfrac{43}{40}\right)\)

\(=1+1+\dfrac{4}{25}.\dfrac{25}{8}\)

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

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

b) \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)

\(=\left(\dfrac{33}{5}-\dfrac{45}{14}\right).\dfrac{35}{6}:\left(-4\right):2,5\)

\(=\left(\dfrac{462}{60}-\dfrac{225}{60}\right).\dfrac{35}{6}.\dfrac{1}{-4}:\dfrac{5}{2}\)

\(=\dfrac{237}{60}.\dfrac{35}{6}.\dfrac{1}{-4}.\dfrac{2}{5}\)

\(=\dfrac{3.79.7.5.2}{5.14.3.2.\left(-4\right).5}\)

\(=\dfrac{79.7}{14.\left(-4\right).5}=\dfrac{553}{-280}\) (số xấu :v)

c) \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)

\(=\dfrac{1}{3}+\dfrac{\dfrac{30}{60}+\dfrac{20}{60}-\dfrac{12}{60}}{\dfrac{45}{60}+\dfrac{30}{60}-\dfrac{9}{60}}\)

\(=\dfrac{1}{3}+\dfrac{\dfrac{19}{30}}{\dfrac{11}{10}}\)

\(=\dfrac{1}{3}+\dfrac{19}{33}\)

\(=\dfrac{11}{33}+\dfrac{19}{33}=\dfrac{30}{33}=\dfrac{10}{11}\)

\(2\dfrac{1}{6}+\dfrac{5}{6}:3-0,125\times\left(-2\right)^2\)

\(=\dfrac{13}{6}+\dfrac{5}{18}-0,125\times4\)

\(=\dfrac{22}{9}-0,5\)

\(=\dfrac{35}{18}\)

\(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)\(=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{60.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}{60.\left(\frac{3}{4}+\frac{1}{2}-\frac{3}{10}\right)}\)

\(=\frac{1}{3}+\frac{30+20-12}{45+30-18}\)\(=\frac{1}{3}+\frac{38}{57}=1\)

\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)

= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{2}{4}+\frac{2}{6}-\frac{2}{10}}{\frac{3}{4}+\frac{3}{6}-\frac{3}{10}}\)

= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{2.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}{3.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}\)

= \(\frac{1}{3}+\frac{2}{3}\)

= \(\frac{3}{3}\)

= 1

=1 nhé bạn

câu trả lơì là 1