Question

ПОМОГИТЕ ПОЖАЛУЙСТА НОМЕР 1458

Answer 1

1458. Розв'язати рівняння:

1.

$\frac{5}{12}x-0,7=2,4-\frac{4}{9}x \\\\\frac{5}{12}x+\frac{4}{9}x= \frac{7}{10}+\frac{12}{5}\\\\ \frac{5\cdot 3+4\cdot 4}{36}x= \frac{7+2\cdot 12}{10} \\\\ \frac{31}{36}x= \frac{31}{10} \\\\ \frac{31}{36}x= \frac{31}{10} \\\\x = \frac{31}{10}:\frac{31}{36} = \frac{36}{10}\cdot \frac{36}{31} = \frac{31}{10} = 3\frac{6}{10}=3,6\\\\x=3,6$

2.

$\frac{1}{8}x-\frac{1}{4}x=\frac{1}{16}x-3\\\\\frac{1}{8}x-\frac{1}{4}x-\frac{1}{16}x=-3\\\\\frac{2-4-1}{16}x=-3\\\\-\frac{3}{16}x=-3 \: | \cdot (-16)\\\\3x=48\\\\x = 16$

3.

$2\frac{1}{2} x-3=-4\frac{1}{2}x-3\frac{1}{3}\\\\\frac{5}{2}x+\frac{9}{2}x=3-3\frac{1}{3}\\\\\frac{14}{2}x = -\frac{1}{3} \\\\7x = -\frac{1}{3} \\\\x=-\frac{1}{3}:7 \\\\x=-\frac{1}{3}\cdot \frac{1}{7} = -\frac{1}{21}\\\\x= -\frac{1}{21}$

4.

$\frac{3}{5}\left(2\frac{7}{9}x+3\frac{1}{3}\right)=\frac{1}{6}x+6\frac{1}{2}\\\\\frac{3}{5}\left(\frac{25}{9}x+\frac{10}{3}\right)=\frac{1}{6}x+\frac{13}{2}\\\\\frac{75}{45}x+\frac{30}{15} = \frac{1}{6}x+\frac{13}{2}\\\\\frac{5}{3}x - \frac{1}{6}x =\frac{13}{2}-2\\\\\frac{10-1}{6}x =\frac{13-4}{2}\\\\\frac{9}{6}x =\frac{9}{2} \\\\\frac{3}{2}x =\frac{9}{2} \\\\x = \frac{9}{2}:\frac{3}{2} = \frac{9}{2}\cdot \frac{2}{3} =\frac{9}{3} = 3 \\\\x=3$