Viết mỗi phân số sau thành tổng các phân số khác nhau và đều có tử số là 1:
a) \(\frac{5}{6}\) b)\(\frac{7}{12}\) c)\(\frac{7}{8}\)
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a)\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
b)\(\frac{7}{12}=\frac{1}{2}+\frac{1}{12}=\frac{1}{3}+\frac{1}{4}\)
c)\(\frac{7}{8}=\frac{1}{8}+\frac{6}{8}\)
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{7}{12}=\frac{1}{3}+\frac{1}{4}\)
a) \(\frac{2}{3} = \frac{4}{6} = \frac{1}{6} + \frac{3}{6} = \frac{1}{6} + \frac{1}{2}\)
b) \(\frac{8}{{15}} = \frac{5}{{15}} + \frac{3}{{15}} = \frac{1}{5} + \frac{1}{3}\)
c) \(\frac{7}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}\)
d) \(\frac{{17}}{{18}} = \frac{9}{{18}} + \frac{6}{{18}} + \frac{2}{{18}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{9}\).
a, \(\dfrac{5}{6}=\dfrac{1}{2}+\dfrac{1}{3}\)
b, \(\dfrac{7}{8}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}\)
c, \(\dfrac{6}{5}=\dfrac{1}{1}+\dfrac{1}{5}\)
d, \(\dfrac{8}{7}=\dfrac{1}{1}+\dfrac{1}{7}\)
\(a,\dfrac{5}{6}=\dfrac{1}{2}+\dfrac{1}{3}\\ b,\dfrac{7}{8}=\dfrac{1}{8}+\dfrac{1}{4}+\dfrac{1}{2}\\ c,\dfrac{6}{5}=\dfrac{1}{1}+\dfrac{1}{5}\\ d,\dfrac{8}{7}=\dfrac{1}{1}+\dfrac{1}{7}\)
a) \(\frac{5}{6}=\frac{1}{3}+\frac{1}{2}\)
b) \(\frac{7}{12}=\frac{1}{12}+\frac{1}{2}\)
c) \(\frac{7}{8}=\frac{1}{2}+\frac{3}{8}\)
\(\frac{5}{6}=\frac{3}{6}+\frac{2}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{7}{12}=\frac{6}{12}+\frac{1}{12}=\frac{1}{2}+\frac{1}{12}\)
\(=\frac{4}{12}+\frac{3}{12}=\frac{1}{3}+\frac{1}{4}\)
\(\frac{7}{8}=\frac{4}{8}+\frac{2}{8}+\frac{1}{8}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)