вправа 4 гдз 7 клас алгебра Мерзляк Полонський

Відповідь: \begin{equation}1)18\frac{5}{12}-\frac{7}{12}·1\frac{19}{21}-\frac{17}{72}·\frac{2}{3}=\end{equation} \begin{equation}=18\frac{5}{12}-\frac{7}{12}·\frac{40}{21}-\frac{17}{2·36}·\frac{2}{3}=\end{equation} \begin{equation}=18\frac{5}{12}-\frac{7}{4·3}·\frac{4·10}{7·3}-\end{equation} \begin{equation}-\frac{17}{2·36}·\frac{2}{3}=\end{equation} \begin{equation}=18\frac{5}{12}-\frac{10}{9}-\frac{17}{108}=\end{equation} \begin{equation}=17\frac{153}{108}-\frac{120}{108}-\frac{17}{108}=\end{equation} \begin{equation}=17\frac{16}{108}=17\frac{4}{27};\end{equation} \begin{equation}2)\left ( 6\frac{3}{4}-5\frac{1}{8}:1\frac{9}{32}\right )·\frac{5}{11}=\end{equation} \begin{equation}=\left ( 6\frac{3}{4}-\frac{41}{8}·\frac{32}{41} \right )·\frac{5}{11}=\end{equation} \begin{equation}=\left ( 6\frac{3}{4}-4 \right )·\frac{5}{11}=\end{equation} \begin{equation}=2\frac{3}{4}·\frac{5}{11}=\frac{11}{4}·\frac{5}{11}=1\frac{1}{4};\end{equation} 3) (-1,42 - (-3,22)) : (-0,4) + (-6) · (-0,7) =
= 1,8 : (-0,4) + 4,2 = -4,5 + 4,2 = -0,3; \begin{equation}4)\left ( -\frac{7}{18}+\frac{11}{12} \right ):\left ( -\frac{19}{48} \right )=\end{equation} \begin{equation}=\left ( -\frac{14}{36}+\frac{33}{36} \right ):\left ( -\frac{19}{48} \right )=\end{equation} \begin{equation}=\frac{19}{36}·\left ( -\frac{48}{19} \right )=\end{equation} \begin{equation}=-\frac{48}{36}=-\frac{4}{3}=-1\frac{1}{3};\end{equation} \begin{equation}5)\left ( -3\frac{1}{12}-2\frac{1}{15} \right ):\left ( -5\frac{3}{20} \right )=\end{equation} \begin{equation}=\left ( -3\frac{5}{60}-2\frac{4}{60} \right ):\left ( -\frac{103}{20} \right )=\end{equation} \begin{equation}=\left ( -5\frac{9}{60} \right )·\left ( -\frac{20}{103} \right )=\end{equation} \begin{equation}=\frac{309}{60}·\frac{20}{103}=\frac{3·103}{3·20}·\frac{20}{103}=1.\end{equation}