\(\dfrac{1}{11}\) * \(\dfrac{27}{46}\) * 1
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\(=\dfrac{34}{11}\cdot\dfrac{23}{17}\cdot\dfrac{27}{46}\cdot\dfrac{22}{9}=\dfrac{34}{17}\cdot\dfrac{23}{11}\cdot\dfrac{22}{46}\cdot\dfrac{27}{9}\)

\(=2\cdot3\cdot\dfrac{1}{2}\cdot2=6\)

20 tháng 7 2018

a) \(\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{18}{60}\) ( MTC: 60)

= \(\dfrac{3.15}{4.15}+\dfrac{3.12}{5.12}-\dfrac{18}{60}\)

= \(\dfrac{45}{60}+\dfrac{36}{60}-\dfrac{18}{60}\)

= \(\dfrac{45+36-18}{60}\)=\(\dfrac{63}{60}=\dfrac{21}{20}\)

b)\(\dfrac{17}{8}-\dfrac{11}{6}-\dfrac{2}{9}\) (MTC:72)

=\(\dfrac{17.9}{8.9}-\dfrac{11.12}{6.12}-\dfrac{2.8}{9.8}\)

= \(\dfrac{153}{72}-\dfrac{132}{72}-\dfrac{16}{72}\)

=\(\dfrac{153-132-16}{72}\)

=\(\dfrac{5}{72}\)

c)\(\dfrac{23}{29}+\dfrac{5}{11}+\dfrac{17}{11}\) (MTC:319)

= \(\dfrac{23.11}{29.11}+\dfrac{5.29}{11.29}+\dfrac{17.29}{11.29}\)

=\(\dfrac{253}{319}+\dfrac{145}{319}+\dfrac{493}{319}\)

=\(\dfrac{253+145+493}{319}\)=\(\dfrac{891}{319}=\dfrac{81}{29}\)

20 tháng 7 2018

c) \(\dfrac{20}{45}+\dfrac{14}{35}+\dfrac{32}{44}\)

= \(\dfrac{4}{9}+\dfrac{2}{5}+\dfrac{8}{11}\)(Rút gọn b/thức)(MTC:495)

=\(\dfrac{4.55}{9.55}+\dfrac{2.99}{5.99}+\dfrac{8.45}{11.45}\)

=\(\dfrac{220}{495}+\dfrac{198}{495}+\dfrac{360}{495}\)

=\(\dfrac{220+198+360}{495}\)=\(\dfrac{778}{495}\)

e)\(17\dfrac{25}{27}+3\dfrac{7}{2}\)

= \(\dfrac{484}{27}+\dfrac{13}{2}\) (MTC:54)

=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)

\(=\dfrac{968}{54}+\dfrac{351}{54}\)

=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)

13 tháng 6 2018

Dấu " / " là phân số nhé

a) 5/-4 . 16/25 + -5/4 . 9/25

= -5/4 . 16/25 + -5/4 . 9/25

= -5/4 . ( 16/25 + 9/25 )

= -5/4 . 1

= -5/4

b) 4 11/23 - 9/14 + 2 12/23 - 5/4

= 103/23 - 9/14 + 58/23 - 5/4

= 103/23 + 58/23 - 9/14 - 5/4

= 7 - 9/14 - 5/4

= 143/28

c) 2 13/27 - 7/15 + 3 14/27 - 8/15

= 67/27 - 7/15 + 95/27 - 8/15

= 67/27 + 95/27 - 7/15 - 8/15

= 6 - 7/15 - 8/15

= 5

12 tháng 3 2017

banhA = \(\dfrac{17}{15}.\dfrac{-31}{125}.\dfrac{1}{2}.\dfrac{10}{17}.\dfrac{-1}{8}\)

= \(\dfrac{17.\left(-31\right).1.10.\left(-1\right)}{15.125.2.17.8}\)

= \(\dfrac{17.\left[\left(-31\right).\left(-1\right)\right].1.2.5}{5.3.125.17.4.2}\)

= \(\dfrac{31.1}{3.125.4}\)

= \(\dfrac{31}{1500}\)

banhB = \(\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)

= \(\left(\dfrac{11}{4}.\dfrac{-9}{9}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(-1\right)\right].\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{-11}{4}.\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{\left(-11\right).4.2}{4.\left(-11\right)\left(-3\right)}\)

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

Ok nhá!

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

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

Bài 1: 

a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)

c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)

17 tháng 4 2018

Violympic toán 6

Đề răng dài thế này thì tui giải từng câu hí

a) \(\dfrac{-7}{9}+\dfrac{5}{12}-\dfrac{13}{18}\left(MSC:36\right)\)

\(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}\)

\(=\dfrac{-13}{36}-\dfrac{26}{36}\)

\(=\dfrac{-39}{36}\)

\(=\dfrac{13}{3}\)

b) \(\dfrac{11}{9}.\dfrac{15}{4}-\dfrac{11}{4}.\dfrac{7}{9}-\dfrac{11}{9}.\dfrac{5}{4}\)

\(=\left(\dfrac{15}{4}-\dfrac{11}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(1-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(\dfrac{4}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\dfrac{-1}{4}.\dfrac{11}{9}\)

\(=\dfrac{-11}{36}\)

14 tháng 4 2018

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)

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

14 tháng 4 2018

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

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

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

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

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

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