a: \(C=\dfrac{-1}{2}:\dfrac{-2}{3}:\dfrac{-3}{4}\)

\(=\dfrac{-1}{2}\cdot\dfrac{3}{-2}\cdot\dfrac{4}{-3}\)

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

b: \(D=\dfrac{-6}{5}:\dfrac{-5}{4}:\dfrac{-4}{3}\)

\(=-\dfrac{6}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\)

\(=\dfrac{-18}{25}\)

c: \(\dfrac{4}{20}+\dfrac{16}{42}+\dfrac{6}{15}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{4}{20}+\dfrac{-3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{16}{42}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}-\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{8}{21}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)

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

d: \(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)

\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{21}+\dfrac{-125}{143}\)

\(=\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Câu d, chỗ \(\dfrac{-21}{21}\) phải sửa thành \(\dfrac{-21}{23}\)

a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)

b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)

c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)

\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)

Bài 1: 

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

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

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

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

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

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

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

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

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

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

a, \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{-3}{5}+\frac{6}{15}+\frac{3}{20}\right)\)\(+\left(\frac{16}{42}+\frac{2}{21}+\frac{-10}{21}\right)\)

\(=\left(\frac{4}{20}+\frac{-12}{20}+\frac{8}{20}+\frac{3}{20}\right)\)\(+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)\)

\(=\frac{3}{20}\)

b,\(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\left(\frac{125}{93}+\frac{-125}{143}\right)\)

\(=\frac{6250}{13299}\)

A.\(\dfrac{4}{20}\)+\(\dfrac{16}{42}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)

=\(\dfrac{1}{5}\)+\(\dfrac{8}{21}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)

=\((\dfrac{1}{5}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)\()\)+\((\dfrac{8}{21}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)\()\)+\(\dfrac{3}{20}\)

=\((\)\(\dfrac{1}{5}\)+\(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)\()\)+ 0 +\(\dfrac{3}{20}\)

= 0 + 0 +\(\dfrac{3}{20}\)

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

B.\(\dfrac{42}{46}\)+\(\dfrac{250}{286}\)+\(\dfrac{-2121}{2323}\)+\(\dfrac{-125125}{143143}\)

=\(\dfrac{21}{23}\)+\(\dfrac{125}{143}\)+\(\dfrac{-21}{23}\)+\(\dfrac{-125}{134}\)

=\((\dfrac{21}{23}+\dfrac{-21}{23})+(\dfrac{125}{143}+\dfrac{-125}{134})\)

= 0 + 0

= 0

Chúc bạn Tran Mai học tốt nha

nick mình đi

Long ơi mình ko biết 

vênh 6G

Vớ va vớ vẩn ấy

Long 6C

Bài 1 câu a:

1/-8 + -5/8

= -1/8 - 5/8

= -6/8

= -3/4

Câu b:

-1/21 + -1/28

= -4/84 - 3/84

= - 7/854

Câu c:

-5/9 + 8/15 + -2/11 + 4/-9 + 7/15

= (8/15+ 7/15) - (5/9 + 4/9) - 2/11

= 1 - 1 - 2/11

= 0 - 2/11

= - 2/11

Câu d:

-25/-75 + 17/34 + 121/- 132

= 1/3 + 1/2 - 11/12

= 4/12 + 6/12 - 11/12

= 10/12 - 11/12

= - 1/12

Bài 2a:

51/23 = 5151/2323 = 515151/232323

\(\frac{5151}{2323}\) = \(\frac{5151:101}{2323:101}\) = \(\frac{51}{23}\) (1)

\(\frac{515151}{232323}\) = \(\frac{515151:10101}{232323:10101}\) = \(\frac{51}{23}\) (2)

Từ (1) và (2) ta có:

\(\frac{51}{23}=\frac{5151}{2323}=\frac{515151}{232323}\) (đpcm)

a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

 b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{17875}{13299}+\frac{-11625}{13299}\)

\(=\frac{6250}{13299}\)

(Hình như câu b hơi sai)

\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

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Câu hỏi của Tran Mai - Toán lớp 6 | Học trực tuyến