помогите пожалуйста . Created by karinakronevald81. matematika-ru

491. Выполните действия:492. Найдите значения выражений:

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karinakronevald81 @karinakronevald81

August 2021 1 4 Report

помогите пожалуйста

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Rakvium

491. Выполните действия:

$1)12m-5\frac{3}{4}m=11m+\frac{4}{4}m-(5m+\frac{3}{4}m)=6\frac{1}{4}\\\\2)8\frac{1}{3}dm-2\frac{1}{4}dm=8\frac{4}{12}dm-2\frac{3}{12}dm=6\frac{1}{12}dm\\\\3)5\frac{3}{4}cm-4\frac{1}{8}cm=5\frac{6}{8}cm-4\frac{1}{8}cm=1\frac{5}{8}cm\\\\4)6h-\frac{2}{7}h=5\frac{5}{7}h\\\\5)3\frac{1}{2}h-2\frac{3}{4}=3h+\frac{2}{4}h-(2h+\frac{3}{4}h)=1h+\frac{2}{4}h-\frac{3}{4}h=\frac{6}{4}h-\frac{3}{4}h=\frac{3}{4}h\\\\6)9h-5\frac{1}{4}h=8h+\frac{4}{4}h-(5h+\frac{1}{4}h)=3\frac{3}{4}h.$

492. Найдите значения выражений:

$1)19\frac{5}{6}-6\frac{1}{12}-2\frac{7}{9}=19\frac{10}{12}-6\frac{1}{12}-2\frac{7}{9}=13\frac{9}{12}-2\frac{7}{9}=13\frac{3}{4}-2\frac{7}{9}=\\\\=11\frac{3}{4}-\frac{7}{9}=11\frac{27}{36}-\frac{28}{36}=11\frac{35}{36}\\\\2)11\frac{7}{10}-5\frac{7}{30}-3\frac{4}{15}=11\frac{21}{30}-5\frac{7}{30}-3\frac{8}{30}=3\frac{6}{30}=3\frac{1}{5}\\\\3)9\frac{14}{15}-2\frac{1}{5}-4\frac{7}{10}=9\frac{28}{30}-2\frac{6}{30}-4\frac{21}{30}=3\frac{1}{30}$

$4)20\frac{3}{8}-(6-2\frac{1}{6})=20\frac{3}{8}-6+2\frac{1}{6}=14\frac{3}{8}+2\frac{1}{6}=\\\\=16+\frac{3}{8}+\frac{1}{6}=16+\frac{3*3}{8*3}+\frac{1*4}{6*4}=16+\frac{9}{24}+\frac{4}{24}=16+\frac{13}{24}=16\frac{13}{24}\\\\5)28\frac{1}{20}-(7-2\frac{8}{15})=28\frac{1}{20}-7+2\frac{8}{15}=23\frac{1}{20}+\frac{8}{15}=\\\\=23\frac{1*3}{20*3}+\frac{8*4}{15*4}=23\frac{3}{60}+\frac{32}{60}=23\frac{35}{60}=23\frac{35:5}{60:5}=23\frac{7}{12}$

$6)29\frac{3}{25}-(4-2\frac{1}{5})=29\frac{3}{25}-(2-\frac{1}{5})=29\frac{3}{25}-2+\frac{1}{5}=27\frac{3}{25}+\frac{1}{5}=\\\\=27\frac{3}{25}+\frac{1*5}{5*5}=27\frac{3}{25}+\frac{5}{25}=27\frac{8}{25}.$

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