a. \(\dfrac{-3}{12}+\dfrac{1}{-4}=\dfrac{-3}{12}+\dfrac{-3}{12}=\dfrac{-3-3}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}\)
 

b. \(\dfrac{5}{12}+\dfrac{-3}{28}=\dfrac{35}{84}+\dfrac{-9}{84}=\dfrac{35+\left(-9\right)}{84}=\dfrac{26}{84}=\dfrac{13}{42}\)
 

c. \(\dfrac{-7}{15}+\dfrac{3}{35}=\dfrac{-49}{105}+\dfrac{9}{105}=\dfrac{-49+9}{105}=\dfrac{-40}{105}=\dfrac{-8}{21}\)
 

d. \(\dfrac{-5}{7}+\dfrac{-3}{4}=\dfrac{-20}{28}+\dfrac{-21}{28}=\dfrac{-20+\left(-21\right)}{28}=\dfrac{-41}{28}\)

a) \(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\)

\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\)

\(=1-\dfrac{1}{31}\)

\(=\dfrac{30}{31}\)

b) \(\dfrac{4}{11\cdot16}+\dfrac{4}{16\cdot21}+...+\dfrac{4}{61\cdot66}\)

\(=\dfrac{4}{5}\cdot\left(\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{61\cdot66}\right)\)

\(=\dfrac{4}{5}\cdot\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)

\(=\dfrac{4}{5}\cdot\left(\dfrac{1}{11}-\dfrac{1}{66}\right)\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{66}\)

\(=\dfrac{4}{66}\)

\(=\dfrac{2}{33}\)

a) A = 5²/(1.6) + 5²/(6.11) + ... + 5²/(26.31)

= 5.[5/(1.6) + 5/(6.11) + ...+ 5/(26.31)]

= 5.(1 - 1/6 + 1/6 - 1/11 + ... + 1/26 - 1/31)

= 5.(1 - 1/31)

= 5.30/31

= 150/31

b) B = 4/(11.16) + 4/(16.21) + ... + 4/(61.66)

= 4/5 .[5/(11.16) + 5/(16.21) + ... + 5/(61.66)]

= 4/5.(1/11 - 1/16 + 1/16 - 1/21 + ... + 1/61 - 1/66)

= 4/5.(1/11 - 1/66)

= 4/5 . 5/66

= 2/33

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)

4/9 x 11 + 4/11 x 13 + 4/13 x 15 + ... + 4/97 x 99

= 2(1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99)

= 2(1/9 - 1/99)

= 20/99

Chúc bạn học tốt!

Bằng 20/99 nhé

a) \(\dfrac{7}{11}+\dfrac{3}{4}+\dfrac{4}{11}+\dfrac{1}{4}=\left(\dfrac{7}{11}+\dfrac{4}{11}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=1+1=2\)

b) \(\dfrac{72}{99}-\dfrac{28}{99}-\dfrac{14}{99}=\dfrac{72-28-14}{99}=\dfrac{30}{99}=\dfrac{10}{33}\)

c) \(69.78+35.97+30.22=\left(69.78+30.22\right)+35.97=100+35.97=135.97\)

Muốn làm những bài toán có chứa hỗn số thì em cần nhớ:

Bước 1: Chuyển các hỗn số có trong phép tính thành phân số.

Bước 2: Thực hiện phép tính theo thứ tự thực hiện phép tính

a, \(\dfrac{8\dfrac{1}{2}}{15:\dfrac{5}{17}}\)

= \(\dfrac{\dfrac{17}{2}}{15\times\dfrac{17}{5}}\)

= \(\dfrac{\dfrac{17}{2}}{3\times17}\)

= \(\dfrac{17}{2}\) x \(\dfrac{1}{3\times17}\)

= \(\dfrac{1}{6}\)

b,\(\dfrac{4\dfrac{4}{5}:\dfrac{4}{17}}{3\dfrac{2}{5}}\)

= \(\dfrac{\dfrac{24}{5}:\dfrac{4}{17}}{\dfrac{17}{5}}\)

= \(\dfrac{\dfrac{24}{5}\times\dfrac{17}{4}}{\dfrac{17}{5}}\)

= \(\dfrac{6\times\dfrac{17}{5}}{\dfrac{17}{5}}\)

= 6

    \(\dfrac{\dfrac{28}{29}:\dfrac{4}{29}}{\dfrac{7}{9}:\dfrac{1}{9}}\)

=    \(\dfrac{\dfrac{28}{29}\times\dfrac{29}{4}}{\dfrac{7}{9}\times\dfrac{9}{1}}\)

= \(\dfrac{7}{7}\)

= 1 

a: \(\dfrac{-1}{2}+\dfrac{2}{3}=\dfrac{-3+4}{6}=\dfrac{1}{6}\)

Số đối là -1/6

b \(-\dfrac{3}{4}+\dfrac{-4}{3}=\dfrac{-9-16}{12}=\dfrac{-25}{12}\)

Số đối là 25/12

c: \(\dfrac{-7}{2}+\dfrac{-3}{4}=\dfrac{-14-3}{4}=\dfrac{-17}{4}\)

Số đối là 17/4

d: \(-2-\dfrac{3}{4}=\dfrac{-8-3}{4}=-\dfrac{11}{4}\)

Số đối là 11/4

1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot10=14\)

2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

\(\dfrac{3}{4}xyz^2+\dfrac{1}{2}xyz^2+\dfrac{-1}{4}xyz^2=\dfrac{3}{4}xyz^2+\dfrac{2}{4}xyz^2-\dfrac{1}{4}xyz^2=xyz^2\)

\(\dfrac{3}{4}xyz^2+\dfrac{1}{2}xyz^2+\dfrac{-1}{4}xyz^2=xyz^2\)

\(\dfrac{3}{4}xyz^2+\dfrac{1}{2}xyz^2+\dfrac{-1}{4}xyz^2\)

= \(\left(\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{-1}{4}\right)xyz^2\)

= xyz²