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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:
$\frac{-9}{15}+\frac{1}{15}=\frac{-8}{15}$
$\frac{-10}{15}+\frac{2}{15}=\frac{-8}{15}$
$\frac{-11}{15}+\frac{3}{15}=\frac{-8}{15}$
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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\(\frac{-2}{15}\) + \(\frac{2}{15}\)
\(\frac{-3}{15}\) + \(\frac{3}{15}\)
\(\frac{-4}{15}\)+ \(\frac{4}{15}\)
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, \(\frac{-7}{12}=\left(\frac{4}{12}-\frac{11}{12}\right)=\left(\frac{1}{12}-\frac{8}{12}\right)=\left(\frac{3}{12}-\frac{10}{12}\right)\)
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: \(\dfrac{-8}{15}=\dfrac{16}{15}-\dfrac{24}{15}\)
\(\dfrac{-8}{15}=\dfrac{22}{15}-\dfrac{30}{15}\)
\(\dfrac{-8}{15}=\dfrac{9}{15}-\dfrac{17}{15}\)
