Alo

=\(\dfrac{5}{13}\).(\(\dfrac{8}{15}\)+\(\dfrac{26}{15}\)-\(\dfrac{8}{15}\))

=\(\dfrac{5}{13}\).\(\dfrac{26}{15}\)

=\(\dfrac{2}{3}\)

\(\dfrac{5}{13}.\dfrac{8}{15}+\dfrac{5}{13}.\dfrac{26}{15}-\dfrac{5}{13}.\dfrac{8}{15}.\\ =\dfrac{5}{13}.\left(\dfrac{8}{15}+\dfrac{26}{15}-\dfrac{8}{15}\right).\\ =\dfrac{5}{13}.\dfrac{26}{15}=\dfrac{2}{3}.\)

\(\dfrac{5}{13}\times\dfrac{8}{15}+\dfrac{5}{13}\times\dfrac{26}{15}-\dfrac{5}{13}\times\dfrac{8}{15}\)

\(=\dfrac{5}{13}\times\left(\dfrac{8}{15}+\dfrac{26}{15}-\dfrac{8}{15}\right)\)

\(=\dfrac{5}{13}\times\dfrac{26}{15}\)

\(\dfrac{2}{3}\)

Ta có: \(B=\dfrac{5}{13}\cdot\dfrac{8}{15}+\dfrac{5}{13}\cdot\dfrac{26}{15}-\dfrac{5}{13}\cdot\dfrac{8}{15}\)

\(=\dfrac{5}{13}\cdot\dfrac{26}{15}\)

\(=\dfrac{130}{195}=\dfrac{2}{3}\)

\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{12}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)

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

\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2.\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)

Bài này khá giống dạng thi của Tú.

mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\)

\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\)

\(=\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\)

\(=1-\dfrac{17}{13}=-\dfrac{4}{13}\)

\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)

\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\left(\dfrac{32}{60}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}.\dfrac{24}{47}\\ =\dfrac{7}{5}+\dfrac{222}{235}\\ =\dfrac{329}{235}+\dfrac{222}{235}\\ =\dfrac{551}{235}\)

\(a,MSC:180\\ \dfrac{17}{12}=\dfrac{17.15}{12.15}=\dfrac{255}{180};\dfrac{31}{18}=\dfrac{31.10}{18.10}=\dfrac{310}{180};\dfrac{8}{15}=\dfrac{8.12}{15.12}=\dfrac{96}{180}\\ b,MSC:75\\ \dfrac{7}{15}=\dfrac{7.5}{15.5}=\dfrac{35}{75};\dfrac{8}{25}=\dfrac{8.3}{25.3}=\dfrac{24}{75};\dfrac{11}{75}=\dfrac{11}{75}\)

a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A =  - 2 + (-1)\\A =  - 3\end{array}\)

b, 

Cách 1:

\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)

Cách 2:

\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B =  \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)