\(\dfrac{-5}{6}.\dfrac{4}{19}+\df...">
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\(=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}-\dfrac{10}{3}\right)\)

\(=\dfrac{4}{19}\cdot\left(\dfrac{-10-7-40}{12}\right)=\dfrac{4}{19}\cdot\dfrac{-57}{12}=-\dfrac{3}{3}=-1\)

2 tháng 4 2023

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

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

=\(-\dfrac{52}{57}\)

f: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=\dfrac{-5}{19}\)

i: \(=\left(\dfrac{9}{24}-\dfrac{18}{24}+\dfrac{14}{24}\right)\cdot\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{5}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}\)

=1/4+1/2=3/4

28 tháng 7 2023

` 7/19 . 8/11 + 3/11 . 7/19 + (-12)/19 `

 

`= 7/19 . ( 8/11 + 3/11 ) + (-12)/19 `

 

`= 7/19 . 11/11 + (-12)/19`

 

`= 7/19 . 1 + (-12)/19 `

 

`= 7/19 + (-12)/19 `

 

`= -5/19 `

 

`( 3/8 + (-3)/4 + 7/12 ) : 5/6 + 1/2`

 

`= 3/8 + (-3)4 + 7/12 . 6/5 + 1/2`

 

`= ( 9+(-18) + 14)/24 . 6/5 + 1/2`

 

`= 5/24 . 6/5 + 1/2`

 

`= 1/4 + 1/2 `

 

`= 3/4`

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

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

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

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

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

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

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

16 tháng 4 2023

1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10

3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3

4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8

5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15

6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11 

7) = 11/12 x -1/8 + 11/12 x -3/16  - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )

8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v ) 

 

18 tháng 7 2023

1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10

3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3

4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8

5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15

6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11 

7) = 11/12 x -1/8 + 11/12 x -3/16  - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )

8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v ) 

\(=\dfrac{-3}{4}-\dfrac{1}{2}-\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{-6}{4}+\dfrac{5}{8}-\dfrac{1}{2}=-2+\dfrac{5}{8}=\dfrac{-16+5}{8}=-\dfrac{11}{8}\)

17 tháng 4 2017

19 tháng 4 2017

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.

Giải bài 76 trang 39 SGK Toán 6 Tập 2 | Giải toán lớp 6

16 tháng 3 2017

tìm 1 cái bằng 0

23 tháng 3 2022

\(\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\)

\(=\dfrac{5}{3}\cdot\left(\dfrac{7.}{25}+\dfrac{21}{25}-\dfrac{7}{25}\right)\)

\(=\dfrac{5}{3}\cdot\dfrac{21}{25}=\dfrac{7}{5}\)

b) \(250\%+19\dfrac{3}{11}\cdot\dfrac{7}{26}-6\dfrac{3}{11}\cdot\dfrac{7}{26}\)

\(=\dfrac{5}{2}+\dfrac{212}{11}\cdot\dfrac{7}{26}-\dfrac{69}{11}\cdot\dfrac{7}{26}\)

\(=\dfrac{7}{26}\cdot\left(\dfrac{212}{11}-\dfrac{69}{11}\right)+\dfrac{5}{2}\)

\(=\dfrac{7}{26}\cdot13+\dfrac{5}{2}\)

\(=\dfrac{7}{2}+\dfrac{5}{2}\)

\(=\dfrac{12}{2}=6\)

 

31 tháng 7 2017

Trời ơi cái đề bài !!!

Thoy thì làm từng câu vậy

a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)

\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\dfrac{7}{222222}\)

\(I=\dfrac{7}{22}\)

31 tháng 7 2017

1.

a) \(A=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)\(A=\dfrac{10101.5}{10101.11}+\dfrac{10101.5}{10101.22}-\dfrac{10101.4}{10101.11}\)

\(A=\dfrac{5}{11}+\dfrac{5}{22}-\dfrac{4}{11}=\dfrac{7}{22}\)

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

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

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

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

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

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