Question

6 класс номер 324d) e)и 325 d) e)
полное решение и
подробно​

Answer 1

$324d)\; \; 21(x-5)=2(10-7x)\\\\21x-105=20-14x\\\\21x+14x=20+105\\\\35x=125\\\\x=\frac{125}{35}\\\\x=\frac{25}{7}=3\frac{4}{7}$

$e)\; \; \frac{2}{1}\, (2x+\frac{1}{3})=2\frac{1}{6}-(\frac{x}{2}-\frac{5}{33})\\\\4x+\frac{2}{3}=\frac{13}{6}-\frac{x}{2}+\frac{5}{33}\; \Big |\cdot 66\\\\264x+44=143-33x+10\\\\297x=109\; \; ,\; \; x=\frac{109}{297}$

$325d)\; \; 12-3(2x+7)=2-9x\\\\12-6x-21=2-9x\\\\3x=11\\\\x=\frac{11}{3}\; \; ,\; \; x=3\frac{2}{3}$

$e)\; \; \frac{1}{3}\, x-\frac{13}{15}=1\frac{1}{2}\, (x+\frac{2}{3})\\\\\frac{1}{3}\, x-\frac{13}{15}=\frac{3}{2}\, x+\frac{3}{2}\cdot \frac{2}{3}\\\\\frac{1}{3}\, x-\frac{3}{2}\, x=\frac{13}{15}+1\\\\\frac{2-9}{6}\, x=\frac{13+15}{15}\\\\-\frac{7}{6}\, x=\frac{28}{15}\\\\x=-\frac{28}{15}\cdot \frac{6}{7}\\\\x=-\frac{4\cdot 2}{5}\; \; ,\; \; x=-\frac{8}{3}\; \; ,\; \; x=-2\frac{2}{3}$