Đề bài yêu cầu gì?

a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)

\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)

\(=\dfrac{15}{26}-\dfrac{4}{26}\)

\(=\dfrac{11}{26}\)

b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)

\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)

\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)

\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)

\(=\dfrac{-967}{63}\)

c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)

\(=-1\)

d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)

=0

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)

ta có

\(\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1=\dfrac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{\sqrt{3}}\)

tương tự ta có

\(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{5}}\)

\(\sqrt{\dfrac{3}{7}}+\sqrt{\dfrac{5}{7}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{7}}\)

\(A=\dfrac{\sqrt{\dfrac{5}{3}}}{\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1}+\dfrac{\sqrt{\dfrac{7}{5}}}{\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{\sqrt{\dfrac{3}{7}}}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{3}{7}}+1}\)

\(A=\dfrac{\sqrt{5}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}+\dfrac{\sqrt{7}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}=1\)

\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)

\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)

\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)

i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

-6/8=-3/2

2/7

Quy đồng được 12/20+-35/20=-23/20

Quy đồng được -10/15+3/15=-7/15

Quy đồng lên được -4/26+-5/26=-9/26

Quy đồng lên được: -12/21+7/21=-5/21 nhé

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

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

\(\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{12}{20}+\dfrac{-35}{20}=\dfrac{-23}{20}\)

\(\dfrac{-2}{3}+\dfrac{1}{5}=\dfrac{-10}{15}+\dfrac{3}{15}=\dfrac{-13}{15}\)

\(\dfrac{2}{13}+\dfrac{-5}{26}=\dfrac{-4}{26}+\dfrac{-5}{26}=\dfrac{-9}{26}\)

\(\dfrac{-4}{7}+\dfrac{1}{3}=\dfrac{-12}{21}+\dfrac{7}{21}=\dfrac{-5}{21}\)

\(A=\dfrac{\dfrac{3}{11}+\dfrac{3}{3}-\dfrac{3}{7}}{\dfrac{9}{11}+\dfrac{9}{3}-\dfrac{9}{7}}-\dfrac{\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{8}}{\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}+\dfrac{7}{16}}\)

\(=\dfrac{1}{3}-1:\dfrac{7}{2}=\dfrac{1}{3}-\dfrac{2}{7}=\dfrac{1}{21}\)

\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)

\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)

\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)

\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)