a)

\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)

b)

\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)

a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)

\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)

\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)

\(=\left[9+1\right]+1+\dfrac{13}{17}\)

\(=10+1+\dfrac{13}{17}\)

\(=11+\dfrac{13}{17}\)

\(=\dfrac{187}{17}+\dfrac{13}{17}\)

\(=\dfrac{200}{17}\)

b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}+\dfrac{12}{7}\)

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

\(=1\)

c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)

\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)

\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)

\(=\left[6+0\right]-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{42}{7}-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)

\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)

\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)

\(=\dfrac{2}{7}.\left[2+0\right]\)

\(=\dfrac{2}{7}.2\)

= \(\dfrac{4}{7}\)

a: \(=\dfrac{32}{9}+\dfrac{13}{6}=\dfrac{32\cdot2+13\cdot3}{18}=\dfrac{64+39}{18}=\dfrac{103}{18}\)

b: \(=\dfrac{43}{8}-\dfrac{43}{6}=\dfrac{-43}{24}\)

c:\(=4-2-\dfrac{1}{6}=2-\dfrac{1}{6}=\dfrac{11}{6}\)

d: \(=5+\dfrac{2}{3}+7+\dfrac{1}{2}-3-\dfrac{1}{2}+1+\dfrac{2}{3}\)

\(=10+\dfrac{4}{3}=\dfrac{34}{3}\)

a ) 

\(\frac{2}{7}+\frac{7}{7}.\frac{14}{25}\)

\(=\frac{2}{7}+1.\frac{14}{25}=\frac{2}{7}+\frac{14}{25}\)

\(=\frac{50}{175}+\frac{98}{175}=\frac{148}{175}\)

b)

\(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{5}{30}-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{1}{6}-\frac{8}{9}\)

\(=\frac{36}{42}+\frac{7}{42}-\frac{8}{9}\)

\(=\frac{43}{42}-\frac{8}{9}=\frac{129}{126}-\frac{112}{126}=\frac{17}{126}\)

tk ủng hộ mk nha!!!!!!!!

a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)

\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)

\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)

\(=9\dfrac{74}{99}-\dfrac{10}{9}\)

\(=\dfrac{965}{99}-\dfrac{10}{9}\)

\(=\dfrac{95}{11}\)

b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)

\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)

\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)

\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)

\(=\dfrac{1275}{22}\)

c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)

\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)

\(=\dfrac{89}{40}\)

d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)

\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)

\(=-\dfrac{25}{77}\)

e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

(câu này chờ mình một chút)

Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!

\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)

B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)

=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)

=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)

=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)

=\(\dfrac{-57}{57}=-1\)

a) A = 3/7

b) B = 73/13

c) C = 37/7

d) D = 12

ba câu a) ,b) ,c) bn đổi ra hỗn số giúp mk nha

tick cho tớ nha

sai câu A với B kìa bạn

a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)

b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)

c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)

\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)

d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)

\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)

a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)

b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)

c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)

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

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

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

\(=4-1\dfrac{3}{4}\)

\(=3\dfrac{3}{4}\)

b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)

\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)

\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)

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

\(=2\dfrac{3}{7}\)