\(\dfrac{3}{4}\), \(\dfrac{1}...">
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27 tháng 7 2017

Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya

e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-22).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))

f)| 5x+21 | = | 2x -63 |

g) -45 - |-3x-96 | - 54=-207

Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)

1 tháng 8 2017

a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)

<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)

<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)

<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)

<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân

a: 2/9=4/18

1/3=6/18

5/18=5/18

b: 7/15=14/30

1/5=6/30

-5/6=-25/30

c: -21/56=-3/7

-3/16=-63/336

5/24=70/336

-21/56=-3/7=-144/336

d: \(\dfrac{-4}{7}=\dfrac{-36}{63}\)

8/9=56/63

\(-\dfrac{10}{21}=-\dfrac{30}{63}\)

e: 3/-20=-3/20=-9/60

-11/-30=11/30=22/60

7/15=28/60

21 tháng 3 2017

2) Tinh nhanh:

a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)

= \(\dfrac{5}{598}\)

21 tháng 3 2017

b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)

= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)

31 tháng 3 2017

a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

31 tháng 3 2017

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước vuiko có tâm trạng mà cười nữalolanglimdim

22 tháng 3 2017

\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{13}{4}-\dfrac{7}{-24}\)

\(=\dfrac{85}{24}\)

\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{-3}{56}-\dfrac{3}{28}\)

\(=\dfrac{-9}{56}\)

\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)

\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)

\(=\dfrac{5}{12}\)

\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)

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

\(=0\)

\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{47}{24}+\dfrac{5}{-8}\)

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

22 tháng 3 2017

Bài 2 : Tính

a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)

\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)

\(=\dfrac{85}{24}\)

b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)

\(=\dfrac{32+\left(-35\right)-6}{56}\)

\(=\dfrac{-9}{56}\)

c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)

\(=\dfrac{7-32+\left(-24\right)}{36}\)

\(=\dfrac{-49}{36}\)

d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\left(-1\right)+1\)

\(=0\)

e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)

\(=2+\left(-1\right)+\dfrac{1}{3}\)

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

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

17 tháng 4 2018

Violympic toán 6

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

CÁCH 1 : A = \(\dfrac{235}{11}-\left(\dfrac{8}{5}+\dfrac{81}{11}\right)\)

A = \(\dfrac{235}{11}-\left(\dfrac{88}{55}+\dfrac{405}{55}\right)\)

A = \(\dfrac{235}{11}-\dfrac{493}{55}\)

A = \(\dfrac{1175}{55}+\dfrac{493}{55}\)

A = \(\dfrac{1668}{55}\)

21 tháng 7 2018

a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)

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

= \(\dfrac{-77}{36}\)

b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)

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

= \(\dfrac{10}{15}+\dfrac{-2}{15}\)

= \(\dfrac{-8}{15}\)

c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)

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

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

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

d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)

= \(\dfrac{1}{2}\) . 11 - 7

= \(\dfrac{11}{2}-\dfrac{14}{2}\)

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

e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)

= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)

= \(\dfrac{3}{4}.-28\)

= \(-21\)

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

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