\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)

\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)

\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)

Bài 1:

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)

A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)

A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))

A = 1 - 1 - \(\frac{228}{210}\)

A = 0 - \(\frac{128}{210}\)

A = - \(\frac{64}{105}\)

Bài 2:

B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)

B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)

B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))

B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))

B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))

B = 3 - \(\frac{97}{65}\)

B = \(\frac{195}{65}\) - \(\frac{97}{65}\)

B = \(\frac{98}{65}\)

\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2=1^{2015}-4^2=1-16=-15\)

\(16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)=\left(16\frac{2}{7}-28\frac{2}{7}\right):\left(\frac{-3}{5}\right)=-12.\frac{-5}{3}=20\)

\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)=-8.\frac{1}{2}:\frac{13}{12}=-8.\frac{1}{2}.\frac{12}{13}=\frac{-48}{13}\)

c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)

\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)

\(\frac{100}{-\frac{100}{41}}=-41\)

a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)

\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)

\(7:-\frac{1}{7}=-49\)