a) \(A=\dfrac{7}{8}\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)+5\dfrac{7}{8}\)

\(A=\dfrac{7}{8}.\left(-1\right)+5\dfrac{7}{8}=5\dfrac{7}{8}-\dfrac{7}{8}=5\).

\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}=\dfrac{-35}{32}\)

a) \(A=\dfrac{-7}{8}.\dfrac{5}{9}-\dfrac{4}{9}.\dfrac{7}{8}+5\dfrac{7}{8}\) 

\(A=\dfrac{7}{8}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)+\dfrac{47}{8}\) 

\(A=\dfrac{7}{8}.-1+\dfrac{47}{8}\) 

\(A=\dfrac{-7}{8}+\dfrac{47}{8}\) 

\(A=5\) 

b) \(B=0,25.1\dfrac{3}{5}.\left(\dfrac{5}{4}\right)^2:\left(\dfrac{-4}{7}\right)\) 

\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}\) 

\(B=\dfrac{-35}{32}\)

\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)

1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)

b:Ta có: \(\dfrac{1}{4}+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

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

\(=-\dfrac{1}{8}-1=-\dfrac{9}{8}\)

\(\text{A=}\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)

\(A=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)

\(A=\left(\dfrac{7}{8}+\dfrac{-1}{4}\right):\dfrac{1}{144}\)

\(A=\dfrac{5}{8}.144=90\)

\(\text{B=}\dfrac{3}{4}.1\dfrac{4}{9}-\left(\dfrac{-3}{4}\right)\)

\(B=\dfrac{3}{4}.\dfrac{13}{9}+\dfrac{3}{4}\)

\(B=\dfrac{13}{12}+\dfrac{3}{4}\)

\(B=\dfrac{11}{6}\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)

b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)

bài này tự almf

a) 119/240

b)1/6

c)2/17

d)-48/13

e)-520/7

a: \(=\dfrac{7}{8}\cdot\dfrac{10+24}{60}=\dfrac{7}{8}\cdot\dfrac{34}{60}=\dfrac{7}{8}\cdot\dfrac{17}{30}=\dfrac{119}{240}\)

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

c: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)-1-\dfrac{15}{17}=1-\dfrac{15}{17}=\dfrac{2}{17}\)

d: \(=\left(-8\right)\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

=-4:(27/12-14/12)

=-4:13/12=-4*12/13=-48/13

e: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)=-\dfrac{520}{7}\)

a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)

b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)

c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)

=-520/7

e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)

=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)

=\(-\dfrac{516}{7}\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)

=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)

=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)

=\(\dfrac{119}{240}\)

b)\(\dfrac{3}{2}-\dfrac{5}{6}:\left(\dfrac{1}{2}\right)^2+\sqrt{4}\)

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

=\(\dfrac{3}{2}-\dfrac{10}{3}+2\)

=\(-\dfrac{11}{6}\) +2

=\(\dfrac{1}{6}\)

\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)

\(=\dfrac{-1}{8}\)

\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{71}{35}\)

\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)

\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)

\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)

\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)

\(=\dfrac{-433}{70}\)

a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)

\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)

\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)

\(=2^3+30-\dfrac{3}{2}\)

\(=36,5\)

b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)

\(=0,1^{-1}-2^2-2^{-4}\)

\(=10-4-\dfrac{1}{16}\)

\(=\dfrac{95}{16}\)

c) \(=3^{3.\dfrac{2}{3}}-\dfrac{1}{\left(-2\right)^2}+\left(\dfrac{27}{8}\right)^{-\dfrac{1}{3}}\)

\(=9-\dfrac{1}{4}+\left(\dfrac{3}{2}\right)^{3.\dfrac{-1}{3}}\)

\(=9-\dfrac{1}{4}+\left(\dfrac{3}{2}\right)^{-1}\)

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

\(=\dfrac{113}{12}\)