Question

Используя сочетательный закон сложения для натуральных чисел, проверьте равенство:а)(3/4+1/6)+7/12=3/4+(1/6+7/12); б)7/15+(2/9+5/6)=(7/15+2/9)+5/6.
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Answer 1

а)(3/4+1/6)+7/12=3/4+(1/6+7/12);

(3/4+1/6)+7/12 = $( \frac{3*3}{3*4} + \frac{2*1}{2*6} )+ \frac{7}{12} = ( \frac{9}{12} + \frac{2}{12} )+ \frac{7}{12} = \frac{11}{12} + \frac{7}{12} = \frac{11+7}{12} = \frac{18}{12} = \frac{3}{2} =1 \frac{1}{2}$

3/4+(1/6+7/12) = $\frac{3}{4} + ( \frac{2*1}{2*6} + \frac{7}{12})= \frac{3}{4} +( \frac{2}{12} + \frac{7}{12} ) = \frac{3}{4} + \frac{9}{12} = \frac{3*3}{3*4} + \frac{9}{12} = \frac{9}{12} + \frac{9}{12} = \frac{18}{12}$=$1 \frac{1}{2}$

$1 \frac{1}{2} =1 \frac{1}{2}$


б)7/15+(2/9+5/6)=(7/15+2/9)+5/6.

7/15+(2/9+5/6)=$\frac{7}{15} +( \frac{2*2}{2*9} + \frac{3*5}{3*6} ) = \frac{7}{15} +( \frac{4}{18} + \frac{15}{18} ) = \frac{7}{15} + \frac{19}{18} = \frac{6*7}{6*15} + \frac{5*19}{5*18} = \frac{42}{90} + \frac{95}{90}$=$\frac{137}{90} =1 \frac{47}{90}$

(7/15+2/9)+5/6 = $( \frac{3*7}{3*15} + \frac{5*2}{5*9} )+ \frac{5}{6} =( \frac{21}{45} + \frac{10}{45} )+ \frac{5}{6} = \frac{31}{45} + \frac{5}{6} = \frac{2*31}{2*45} + \frac{15*5}{15*6} = \frac{62}{90} + \frac{75}{90}$=$\frac{137}{90} =1 \frac{47}{90}$
$1 \frac{47}{90} =1 \frac{47}{90}$