Question

решите срочно пожалуйста

Answer 1

Ответ:

$1)1-\frac{4}{9} =\frac{9}{9}- \frac{4}{9}=\frac{5}{9} \\2)1+1\frac{5}{6} =\frac{6}{6}+\frac{11}{6}=\frac{17}{6}=2\frac{5}{6} \\3)7-\frac{3}{8}=\frac{56}{8} -\frac{3}{8}=\frac{53}{8}=6\frac{5}{8} \\4)4-3\frac{3}{4}=\frac{16}{4}-\frac{15}{4}=\frac{1}{4}\\5) 20-(3\frac{1}{6} +2\frac{3}{4})=20-(\frac{19}{6}+\frac{11}{4})=20-(\frac{38}{12}+\frac{33}{12})=20-\frac{71}{12}=\frac{240}{12}-\frac{71}{12}=\frac{169}{12}= 14\frac{1}{12}$

$6)1\frac{5}{6} +( 3-\frac{7}{10})=1\frac{5}{6}+(\frac{30}{10}-\frac{7}{10})=\frac{11}{6}+\frac{23}{10}=\frac{55}{30}+ \frac{69}{30}=\frac{124}{30} =4\frac{4}{30}=4\frac{2}{15}$

Пошаговое объяснение:

Answer 2

$1)1-\dfrac{4}{9} =\dfrac{9}{9}-\dfrac{4}{9}=\boxed{\dfrac{5}{9}}\\\\2)1+1\dfrac{5}{6}=1+1+\dfrac{5}{6}=\boxed{2\dfrac{5}{6}}\\\\3)7-\dfrac{3}{8}=6+\Big(1-\dfrac{3}{8}\Big)=6+\Big(\dfrac{8}{8}-\dfrac{3}{8}\Big)=6+\dfrac{5}{8}=\boxed{6\dfrac{5}{8}}\\\\4)4-3\dfrac{3}{4}=4-3-\dfrac{3}{4}=1-\dfrac{3}{4}=\dfrac{4}{4}-\dfrac{3}{4}=\boxed{\dfrac{1}{4}}\\\\5)20-\Big(3\dfrac{1}{6}+2\dfrac{3}{4}\Big)=20-\Big(3+2+\dfrac{1}{6}+\dfrac{3}{4}\Big)=20-\Big(5+\dfrac{2+9}{12}\Big)=$

$=20-5-\dfrac{11}{12}=15-\dfrac{11}{12}=14+\Big(1-\dfrac{11}{12}\Big)=14+\dfrac{1}{12}=\boxed{14\dfrac{1}{12} }\\\\6)1\dfrac{5}{6} +\Big(3-\dfrac{7}{10}\Big)=1\dfrac{5}{6}+\Big(2+1-\dfrac{7}{10}\Big)=1\dfrac{5}{6} +\Big(2+\dfrac{10}{10}-\dfrac{7}{10} \Big)=\\\\=1\dfrac{5}{6}+2\dfrac{3}{10}=1+2+\dfrac{5}{6}+\dfrac{3}{10}=3+\dfrac{25+9}{30} =3+\dfrac{34}{30}=\\\\=3+1\dfrac{4}{30}=3+1\dfrac{2}{15} =\boxed{4\dfrac{2}{15}}$