(49 5/24-46 7/20):2 1/3+0,6

Решите задачу:

$(49 \frac{5}{24} -46 \frac{7}{20} ):2 \frac{1}{3} +0,6= 1 \frac{33}{40} \\ \\ 1)~~49 \frac{5}{24} -46 \frac{7}{20} =49-46+ \frac{5}{24}-\frac{7}{20}=3+ \frac{5*5}{24*5} -\frac{7*6}{20*6}= \\ =3+ \frac{25}{120} -\frac{42}{120} =3\frac{25}{120} -\frac{42}{120} =2\frac{25+120}{120} -\frac{42}{120} =2\frac{145}{120} -\frac{42}{120} = \\ =2\frac{145-42}{120}=2\frac{103}{120} \\ \\$

$\\ \\2)~~2\frac{103}{120} :2 \frac{1}{3} =\frac{120*2+103}{120} : \frac{3*2+1}{3} = \frac{343}{120} : \frac{7}{3}=\frac{343}{120} * \frac{3}{7}= \\ =\frac{49*7}{40*3} * \frac{3}{7}= \frac{49}{40}= \frac{40+9}{40}=1 \frac{9}{40} \\ \\ 3)~~1 \frac{9}{40}+0.6=1 \frac{9}{40}+ \frac{6}{10} =1\frac{9}{40}+ \frac{6*4}{10*4}=1 \frac{9}{40}+ \frac{24}{40} =1 \frac{9+24}{40}=1 \frac{33}{40}$