Question

решите уравнения:а)8/25у=3цел.1/5 б)3цел.1/12 -1цел.5/6n=1цел.17/24 в) (7/12+11/30х):7цел.1/4=1/3 г)1/9z+7/18z-11/27z=2цел.1/2 д

) а-13/18а=1/3

Answer 1

а. $\frac{8}{25}y=3\frac{1}{5}$

$\frac{8}{25}y=\frac{16}{5} \\ y=\frac{16}{5}:\frac{8}{25} \\ y=10$

б. 3$3\frac{1}{12}- 1\frac{5}{6}n=1\frac{17}{24}\\-1\frac{5}{6}n=1\frac{17}{24}-3\frac{1}{12}\\-\frac{11}{6}n=\frac{41}{24}-\frac{37}{12}\\-\frac{11}{6}n=\frac{41}{24}-\frac{74}{24}\\-\frac{11}{6}n=-\frac{33}{24}\\n=-\frac{33}{24}:-\frac{11}{6}\\ n=\frac{3}{4}$

в.$(\frac{7}{12}+\frac{11}{30}x):7\frac{1}{4}=\frac{1}{3}\\\frac{7}{12}+\frac{11}{30}x=\frac{29}{4}*\frac{1}{3}\\\frac{7}{12}+\frac{11}{30}x=\frac{29}{12}\\\frac{11}{30}x=\frac{29}{12}-\frac{7}{12}\\\frac{11}{30}x=\frac{22}{12}\\x=\frac{22}{12}:\frac{11}{30}\\x=5$

г. $\frac{1}{9}z+\frac{7}{18}z-\frac{11}{27}z=2\frac{1}{2}\\z(\frac{1}{9}=\frac{7}{18}-\frac{11}{27})=\frac{5}{2}\\z(\frac{2}{18}+\frac{7}{18}-\frac{11}{27})=\frac{5}{2}\\z(\frac{9}{18}-\frac{11}{27})=\frac{5}{2}\\z(\frac{27}{54}-\frac{22}{54})=\frac{5}{2}\\z\frac{5}{54}=\frac{5}{2}\\z=\frac{5}{2}:\frac{5}{24}\\z=12$

д.$a-\frac{13}{18}a=\frac{1}{3}\\a(1-\frac{13}{18})=\frac{1}{3}\\a(\frac{18}{18}-\frac{13}{18})=\frac{1}{3}\\a\frac{5}{18}=\frac{1}{3}\\a=\frac{1}{3}:\frac{5}{18}\\a=\frac{6}{5}=1\frac{1}{5}$