Question

Помогите с алгеброй! Пожалуйста , 40б

Answer 1

В первом действительно опечатка. Дробь уж очень смешная..))

$\displaystyle \tt a). \ \ \bigg(26\frac{2}{3}:6,4\bigg)\cdot(19,2:3,(5))-\frac{8\dfrac{4}{7}:2\dfrac{26}{77}}{0,5-18\dfrac{2}{3}\cdot11}-\frac{1}{18}=22\frac{5114}{11061};\\\\\\3,(5)=3\frac{5}{9};\\\\1). \ \ 26\frac{2}{3}:6,4=\frac{80}{3}\cdot\frac{10}{64}=4\frac{1}{6};\\\\2). \ \ 19,2:3,(5)=19,2:3\frac{5}{9}=\frac{192}{10}\cdot\frac{9}{32}=5,4;\\\\3). \ \ 4\frac{1}{6}\cdot5,4=\frac{25}{6}\cdot\frac{54}{10}=22,5;\\\\4). \ \ 8\frac{4}{7}:2\frac{26}{77}=\frac{60}{7}\cdot\frac{77}{180}=\frac{11}{3}=3\frac{2}{3};$

$\displaystyle \tt 5). \ \ 18\frac{2}{3}\cdot11=\frac{56\cdot11}{3}=205\frac{1}{3};\\\\6). \ \ 0,5-205\frac{1}{3}=-204\frac{5}{6};\\\\7). \ \ 3\frac{2}{3}:\bigg(-204\frac{5}{6}\bigg)=-\frac{11}{3}\cdot\frac{6}{1229}=-\frac{22}{1229};\\\\8). \ \ 22,5-\bigg(-\frac{22}{1229}\bigg)=\frac{55305+44}{2458}=\frac{55349}{2458}=22\frac{1273}{2458};$

$\displaystyle \tt 9). \ \ 22\frac{1273}{2458}-\frac{1}{18}=\frac{18\cdot55349-2458}{2458\cdot18}= \frac{496912}{22122}=22\frac{5114}{11061};$

$\displaystyle \tt b).\frac{9x+3y}{2x+2,(6)y}=\frac{3\cdot(3x+y)}{2x+2\dfrac{2}{3}y};\\\\\\2,(6)=2\frac{2}{3}};\\\\npu \ \ \ 3x=5y\\\\{} \ \ \ \ \ \ \ \ x=\frac{5}{3}y\\\\\frac{3\cdot(5y+y)}{\dfrac{10}{3}y +\dfrac{8}{3}y}=\frac{3\cdot6y}{6y}=3;$