Question

№1394.Найдите значение $\frac{x}{y}$ ,если:
$а)x = \frac{7}{15} + \frac{7}{30} + \frac{4}{5} , \: а \: y = 2 - \frac{1}{3} - \frac{1}{2}$

$б)x = \frac{1}{6} + \frac{1}{10} + \frac{1}{15} , \: \: а \: \: y = \frac{3}{5} - \frac{1}{3} - \frac{1}{4}$
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Answer 1

1)

$x=\frac{7}{15} +\frac{7}{30} +\frac{4}{5} =\frac{14}{30} +\frac{7}{30} +\frac{24}{30} =\frac{45}{30} =\frac{3}{2}$

$y=2-\frac{1}{3} -\frac{1}{2} =\frac{12}{6} -\frac{2}{6} -\frac{3}{6}=\frac{7}{6}$

$\frac{x}{y} =\frac{3}{2} :\frac{7}{6} =\frac{3}{2} *\frac{6}{7} =\frac{9}{7} =1\frac{2}{7}$

2)

$x=\frac{1}{6} +\frac{1}{10} +\frac{1}{15} =\frac{5}{30} +\frac{3}{30} +\frac{2}{30} =\frac{10}{30} =\frac{1}{3}$

$y=\frac{3}{5} -\frac{1}{3} -\frac{1}{4} =\frac{36}{60} -\frac{20}{60} -\frac{15}{60} =\frac{1}{60}$

$\frac{x}{y} =\frac{1}{3} :\frac{1}{60} =\frac{1}{3} *60=20$