\(\dfrac{423}{2813}\)-\(\dfrac{5}{6}\)=

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29 tháng 6 2018

Giải:

\(\dfrac{423}{2813}-\dfrac{5}{6}\)

\(=0,150373267-0,8\left(3\right)\)

\(=-6829600664\)

Vậy ...

\(\dfrac{423}{2813}-\dfrac{5}{6}=\dfrac{2538}{16878}-\dfrac{14065}{16878}=\dfrac{-11257}{16878}\)

29 tháng 6 2018

\(\dfrac{423}{2813}-\dfrac{5}{6}=\dfrac{2538}{16878}+\dfrac{14065}{16878}=\dfrac{16603}{16878}\)

30 tháng 3 2018

các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho

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

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

22 tháng 4 2018

giúp mình đi mà ToT khocroi

17 tháng 4 2018

Violympic toán 6

4 tháng 8 2017

Bài 1: Tính ( hợp lý nếu có thể )

\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)

\(=-1+1+\dfrac{2}{-5}\)

\(=0+\dfrac{2}{-5}\)

\(=\dfrac{2}{-5}\)

\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)

\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}\)

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

\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1\)

\(=0\)

\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)

\(=\dfrac{1}{6}+\dfrac{-1}{6}\)

\(=0\)

4 tháng 8 2017

Bài 2: Tìm x,biết:

a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}-\dfrac{2}{3}\)

\(x=\dfrac{2}{15}\)

Vậy \(x=\dfrac{2}{15}\)

b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}+\dfrac{2}{3}\)

\(x=\dfrac{3}{3}=1\)

Vậy \(x=1\)

c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!

\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)

\(x=\dfrac{1}{44}\)

Vậy \(x=\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\)

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

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

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

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

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

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

26 tháng 3 2018

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

30 tháng 3 2018

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: \(\Leftrightarrow x\cdot\dfrac{1}{2}-\dfrac{3}{5}x+\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7}{10}x\)

=>3/5x=-4

hay x=-4:3/5=-20/3

b: \(\Leftrightarrow4x-6-9=5-3x-2\)

=>4x-15=-3x+3

=>7x=18

hay x=18/7