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a) Ta có BCNN(3,7)=21
Thừa số phụ: 21:3=7 và 21:7=3
\(\dfrac{2}{3} = \dfrac{{2.7}}{{3.7}} = \dfrac{{14}}{{21}}\) và \(\dfrac{{ - 6}}{7} = \dfrac{{ - 6.3}}{{7.3}} = \dfrac{{ - 18}}{{21}}\)
b) Ta có \(BCNN\left( {\left( {{2^2}{{.3}^2}} \right),\left( {{2^2}.3} \right)} \right) = {2^2}{.3^2}\)
Thừa số phụ \(\left( {{2^2}{{.3}^2}} \right):\left( {{2^2}.3^2} \right) = 1\) và \(\left( {{2^2}{{.3}^2}} \right):\left( {{2^2}.3} \right) = 3\)
\(\dfrac{5}{{{2^2}{{.3}^2}}}\) và \(\dfrac{{ - 7}}{{{2^2}.3}} = \dfrac{{ - 7.3}}{{{2^2}{{.3}^2}}} = \dfrac{{ - 21}}{{{2^2}{{.3}^2}}}\)
a: \(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{9}{3}\)
\(\dfrac{7}{3}+\dfrac{2}{3}=\dfrac{9}{3}\)
=>\(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{7}{3}+\dfrac{2}{3}\)
\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}\)
=>\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{4}{5}+\dfrac{3}{5}\)
b: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{7+16}{9}=\dfrac{23}{9}\)
\(\dfrac{16}{9}+\dfrac{7}{9}=\dfrac{16+7}{9}=\dfrac{23}{9}\)
Do đó: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{16}{9}+\dfrac{7}{9}\)
a) \(\dfrac{2}{3}=\dfrac{2.8}{3.8}=\dfrac{16}{24}\)
\(\dfrac{5}{8}=\dfrac{5.3}{8.3}=\dfrac{15}{24}\)
b) \(\dfrac{1}{4}=\dfrac{1.3}{4.3}=\dfrac{3}{12}\)
\(\dfrac{7}{12}=\dfrac{7.1}{12.1}=\dfrac{7}{12}\)
c) \(\dfrac{5}{6}=\dfrac{5.4}{6.4}=\dfrac{20}{24}\)
\(\dfrac{3}{8}=\dfrac{3.3}{8.3}=\dfrac{9}{24}\)
a) 23=2.83.8=162423=2.83.8=1624
58=5.38.3=152458=5.38.3=1524
b) 14=1.34.3=31214=1.34.3=312
712=7.112.1=712712=7.112.1=712
c) 56=5.46.4=202456=5.46.4=2024
38=3.38.3=924
