Câu 1.

$\frac{1}{15}-\frac{9}{15}=\frac{-8}{15}$

$\frac{2}{15}-\frac{10}{15}=\frac{-8}{15}$

$\frac{3}{15}-\frac{11}{15}=\frac{-8}{15}$

Câu 2:

$\frac{-9}{15}+\frac{1}{15}=\frac{-8}{15}$

$\frac{-10}{15}+\frac{2}{15}=\frac{-8}{15}$

$\frac{-11}{15}+\frac{3}{15}=\frac{-8}{15}$

Câu 3:

$\frac{-7}{15}-\frac{1}{15}=\frac{-8}{15}$

$\frac{-6}{15}-\frac{2}{15}=\frac{-8}{15}$

$\frac{-5}{15}-\frac{3}{15}=\frac{-8}{15}$

\(\frac{-2}{15}\) + \(\frac{2}{15}\)

\(\frac{-3}{15}\) + \(\frac{3}{15}\)

\(\frac{-4}{15}\)+ \(\frac{4}{15}\)

1/ Cách 1: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-1+\left(-7\right)}{15}\) = \(\dfrac{-1}{15}\) + \(\dfrac{-7}{15}\)

Cách 2: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-2+\left(-6\right)}{15}\) = \(\dfrac{-2}{15}\) + \(\dfrac{-6}{15}\) = \(\dfrac{-2}{15}\) + \(\dfrac{-2}{5}\)

Cách 3: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-3+\left(-5\right)}{15}\) = \(\dfrac{-3}{15}\) + \(\dfrac{-5}{15}\) = \(\dfrac{-1}{5}\) + \(\dfrac{-1}{3}\)

2/ C1: Ta có: \(\dfrac{-8}{15}=\dfrac{10-18}{15}=\dfrac{10}{15}-\dfrac{18}{15}=\dfrac{2}{3}-\dfrac{6}{5}\)

C2: Ta có: \(\dfrac{-8}{15}=\dfrac{1-9}{15}=\dfrac{1}{15}-\dfrac{9}{15}=\dfrac{1}{15}-\dfrac{3}{5}\)

C3: Ta có: \(\dfrac{-8}{15}=\dfrac{5-13}{15}=\dfrac{5}{15}-\dfrac{13}{15}=\dfrac{1}{3}-\dfrac{13}{15}\)

3/C1: Ta có: \(\dfrac{-8}{15}=\dfrac{-16+8}{15}=\dfrac{-16}{15}+\dfrac{8}{15}\)

C2: Ta có: \(\dfrac{-8}{15}=\dfrac{-20+12}{15}=\dfrac{-20}{15}+\dfrac{12}{15}=\dfrac{-4}{3}+\dfrac{4}{5}\)

C3: Ta có:\(\dfrac{-8}{15}=\dfrac{-14+6}{15}=\dfrac{-14}{15}+\dfrac{6}{15}=\dfrac{-14}{15}+\dfrac{2}{5}\)

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a: \(\dfrac{-8}{15}=\dfrac{-2}{15}+\dfrac{-6}{15}\)

\(\dfrac{-8}{15}=\dfrac{-1}{15}+\dfrac{-7}{15}\)

\(\dfrac{-8}{15}=\dfrac{-4}{15}+\dfrac{-4}{15}\)

b: \(-\dfrac{8}{15}=\dfrac{17}{15}-\dfrac{25}{15}\)

\(\dfrac{-8}{15}=\dfrac{9}{15}-\dfrac{17}{15}\)

\(\dfrac{-8}{15}=\dfrac{10}{15}-\dfrac{18}{15}\)

a) \(-\dfrac{7}{13}=\dfrac{-3}{13}+\dfrac{-4}{13}\)

b) \(-\dfrac{11}{25}=\dfrac{6}{25}-\dfrac{17}{25}\)

a) Cách 1 : \(-\dfrac{7}{13}=\dfrac{-6-7}{13}=-\dfrac{6}{13}-\dfrac{7}{13}\)

Cách 2 : \(-\dfrac{7}{13}=-\dfrac{1}{13}-\dfrac{10}{13}\)

b) Cách 1 : \(-\dfrac{11}{25}=\dfrac{10}{25}-\dfrac{21}{25}\)

Cách 2 : \(-\dfrac{11}{25}=\dfrac{9}{25}-\dfrac{22}{25}\)

a, - \(\dfrac{11}{25}\) = \(\dfrac{-6}{25}\) + \(\dfrac{-5}{25}\) + 

    - \(\dfrac{11}{25}\) =  \(\dfrac{-1}{25}\) + \(\dfrac{-10}{25}\) + 

    - \(\dfrac{11}{25}\) = \(\dfrac{-3}{25}\) + \(\dfrac{-8}{25}\) +  

b, - \(\dfrac{11}{25}\) = \(\dfrac{6}{25}\) - \(\dfrac{17}{25}\)

   - \(\dfrac{11}{25}\) = \(\dfrac{7}{25}\) - \(\dfrac{18}{25}\)

  - \(\dfrac{11}{25}\)  = \(\dfrac{8}{25}\) - \(\dfrac{19}{25}\)

a)\(\dfrac{-3}{15}+\dfrac{-5}{15}\)

b)\(\dfrac{4}{3}-\dfrac{4}{5}\)

c)\(-1+\dfrac{7}{15}\)

d)\(\dfrac{-1}{15}-\dfrac{7}{15}\)

a: \(\dfrac{-8}{15}=\dfrac{-6}{15}+\dfrac{-2}{15}\)

\(\dfrac{-8}{15}=\dfrac{-3}{15}+\dfrac{-5}{15}\)

\(\dfrac{-8}{15}=\dfrac{-1}{15}+\dfrac{-7}{15}\)

b: \(\dfrac{-8}{15}=\dfrac{16}{15}-\dfrac{24}{15}\)

\(\dfrac{-8}{15}=\dfrac{22}{15}-\dfrac{30}{15}\)

\(\dfrac{-8}{15}=\dfrac{9}{15}-\dfrac{17}{15}\)

1. a, \(\frac{-4}{15}=\left(\frac{-1}{15}+\frac{-3}{15}\right)=\left(\frac{-2}{15}+\frac{-2}{15}\right)=\left(\frac{-0}{15}+\frac{4}{15}\right)\)

b, \(\frac{-4}{15}=\left(\frac{4}{15}-\frac{8}{15}\right)=\left(\frac{3}{15}-\frac{7}{15}\right)=\left(\frac{5}{15}-\frac{9}{15}\right)\)

2 . \(\frac{-7}{12}=\left(\frac{-1}{12}+\frac{-1}{2}\right)=\left(\frac{-1}{6}+\frac{-5}{12}\right)=\left(\frac{-1}{4}+\frac{-1}{3}\right)\)

b, \(\frac{-7}{12}=\left(\frac{4}{12}-\frac{11}{12}\right)=\left(\frac{1}{12}-\frac{8}{12}\right)=\left(\frac{3}{12}-\frac{10}{12}\right)\)