\(a.\)

\(\dfrac{3}{10}:\left(-\dfrac{2}{3}\right)=\dfrac{3}{10}\cdot\dfrac{-3}{2}=-\dfrac{9}{20}\)

\(b.\)

\(\left(-\dfrac{7}{12}\right):\left(-\dfrac{5}{6}\right)=\left(-\dfrac{7}{12}\right)\cdot\left(-\dfrac{6}{5}\right)=\dfrac{\left(-7\right)\cdot\left(-6\right)}{12\cdot5}=\dfrac{7}{10}\)

\(c.\)

\(\left(-15\right):-\dfrac{9}{10}=\left(-15\right)\cdot-\dfrac{10}{9}=\dfrac{150}{9}=\dfrac{50}{3}\)

a) \(\dfrac{3}{10}:\dfrac{-2}{3}=\dfrac{3}{10}.\dfrac{-3}{2}=\dfrac{3.-3}{10.2}=\dfrac{-9}{20}\) 

b) \(\dfrac{-7}{12}:\dfrac{-5}{6}=\dfrac{-7}{12}.\dfrac{-6}{5}=\dfrac{-7.-6}{12.5}=\dfrac{7}{10}\) 

c)\(-15:\dfrac{-9}{10}=-15.\dfrac{-10}{9}=\dfrac{-15.-10}{9}=\dfrac{50}{3}\)

a, 3/5 : -5 = -3/25

b, 24 : -6/7 = -28

c, -4/15 : 2 = -2/15

a, 3/5 : -5 = -3/25

b, 24 : -6/7 = -28

c, -4/15 : 2 = -2/15

\(\dfrac{14}{21}=\dfrac{14:7}{21:7}=\dfrac{2}{3}=\dfrac{1}{3}+\dfrac{1}{3}\)

\(\dfrac{15}{29}=\dfrac{1+14}{29}=\dfrac{1}{29}+\dfrac{14}{29}\)

\(\dfrac{9}{35}=\dfrac{1+8}{35}=\dfrac{1}{35}+\dfrac{8}{35}\)

\(\dfrac{16}{27}=\dfrac{2+14}{27}=\dfrac{2}{27}+\dfrac{14}{27}\)

thank

Giải:

a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\) 

\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\) 

\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\) 

\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\) 

\(=\dfrac{1}{2^6}=\dfrac{1}{64}\) 

b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\) 

\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\) 

\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\) 

\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\) 

\(=2\) 

c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\) 

\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\) 

\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\) 

\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\) 

\(=\dfrac{2.3}{3.5}\) 

\(=\dfrac{2}{5}\) 

Chúc bạn học tốt!

\(\dfrac{11}{16}\) = \(\dfrac{1}{16}\) + \(\dfrac{10}{16}\) = \(\dfrac{1}{16}\) + \(\dfrac{2+8}{2\times8}\) = \(\dfrac{1}{16}\) + \(\dfrac{2}{2\times8}\) + \(\dfrac{8}{2\times8}\) = \(\dfrac{1}{16}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{2}\)

\(\dfrac{5}{12}\) = \(\dfrac{1}{12}\) + \(\dfrac{4}{12}\) = \(\dfrac{1}{12}\) + \(\dfrac{4}{3\times4}\) = \(\dfrac{1}{12}\) + \(\dfrac{1}{3}\)

\(\dfrac{6}{35}\) = \(\dfrac{1}{35}\) + \(\dfrac{5}{35}\) = \(\dfrac{1}{35}\) + \(\dfrac{5}{5\times7}\) = \(\dfrac{1}{35}\) + \(\dfrac{1}{7}\) 

a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)

\(\dfrac{3}{21}\)

\(\dfrac{7}{25}\)

Chọn C

a) \(\dfrac{8}{16}=\dfrac{8:8}{16:8}=\dfrac{1}{2}\)

b) \(\dfrac{10}{30}=\dfrac{10:10}{30:10}=\dfrac{1}{3}\)

c) \(\dfrac{24}{18}=\dfrac{24:6}{18:6}=\dfrac{4}{3}\)

d) \(\dfrac{20}{28}=\dfrac{20:4}{28:4}=\dfrac{5}{7}\)

a)\(\dfrac{56}{70}=\dfrac{56:14}{70:14}=\dfrac{4}{5}\)

b)\(\dfrac{-120}{86}=\dfrac{-120:2}{86:2}=-\dfrac{60}{43}\)

c)\(\dfrac{915+85}{1017-117}=\dfrac{1000}{900}=\dfrac{10}{9}\)

d)\(\dfrac{3838}{5454}=\dfrac{19}{27}\)

a) \(\dfrac{56}{70}=\dfrac{4}{5}\)

\(b,-\dfrac{120}{86}=-\dfrac{60}{43}\)

\(c,\dfrac{10}{9}\)

\(d,\dfrac{19}{27}\)