Question

СРОЧНО! ДАЮ 25 БАЛЛОВ! #1329
Знайдіть корінь рівняння:

Answer 1

1)$\frac{3}{5}(\frac{x}{2}+\frac{2}{3})=\frac{9}{4}+\frac{53x}{10}\\\frac{3x}{10}+\frac{2}{5}=\frac{9}{4}+\frac{53x}{10}\\-\frac{9}{4}+\frac{2}{5}=\frac{53x}{10}-\frac{3x}{10}\\-\frac{37}{20}=5x\\x=-\frac{37}{100}=-0.37$

2)$5-y=8-\frac{1}{3}(\frac{9y}{2}-5)\\5-y=8-\frac{3y}{2}+\frac{5}{3}\\-y+\frac{3y}{2}=8+\frac{5}{3}-5\\\frac{y}{2}=\frac{14}{3}\\y=\frac{28}{3}$

3)$\frac{1}{2}(x-4)+6x=5-\frac{3x}{2}\\\frac{x}{2}-2+6x=5-\frac{3x}{2}\\\frac{x}{2}+6x+\frac{3x}{2}=5+2\\8x=7\\x=\frac{7}{8}$

4)$\frac{16}{5}(1-2y)=\frac{7}{10}(3y-\frac{3}{2})\\\frac{16}{5}-\frac{32y}{5}=\frac{21y}{10}-\frac{21}{20}\\-\frac{32y}{5}-\frac{21y}{10}=-\frac{21}{20}-\frac{16}{5}\\\frac{17y}{2}=\frac{17}{4}\\y=\frac{1}{2}$

5)$\frac{5}{12}(z-3)=\frac{1}{6}(2z-7)+2\\\frac{5z}{12}-\frac{15}{12}=\frac{z}{3}-\frac{7}{6}+2\\\frac{5z}{12}-\frac{z}{3}=-\frac{7}{6}+2+\frac{15}{12}\\\frac{z}{12}=\frac{25}{12}\\z=25$

6)$\frac{5}{8}(x-2)=\frac{2}{3}(x+2)-(3-x)\\\frac{5x}{8}-\frac{5}{4}=\frac{2x}{3}+\frac{4}{3}-3+x\\\frac{5x}{8}-\frac{2x}{3}-x=\frac{4}{3}-3+\frac{5}{4}\\-\frac{25x}{24}=-\frac{5}{12}\\x=\frac{2}{5}$

7)$\frac{y}{6}-(\frac{1}{2}+\frac{8y}{9})=\frac{y}{9}-(\frac{1}{3}+y)\\\frac{y}{6}+\frac{1}{2}-\frac{8y}{9}=\frac{y}{9}+\frac{1}{3}-y\\\frac{y}{6}-\frac{8y}{9}-\frac{y}{9}+y=\frac{1}{3}-\frac{1}{2}\\\frac{y}{6}=-\frac{1}{6}\\y=-1$

8)$\frac{3}{5}(3-2z)=\frac{2}{5}(9-z)-\frac{3}{10}(z-9)\\\frac{9}{5}-\frac{6z}{5}=\frac{18}{5}-\frac{2z}{5}-\frac{3z}{10}+\frac{27}{10}\\-\frac{6z}{5}+\frac{2z}{5}+\frac{3z}{10}=\frac{18}{5}-\frac{9}{5}+\frac{27}{10}\\-\frac{z}{2}=\frac{9}{2}\\z=-9$