§ 3. Упражнение 595. «Математика. 6 класс. Учебник для общеобразовательных организаций. В 2 частях. Часть 1»

    Упражнение 595

    Выполните действия:

      • $$1)\ (7{,}061:2{,}3-2{,}2)\cdot(4{,}2+17{,}391:5{,}27);$$$$2)\ (3{,}7+14{,}058:6{,}39)\cdot(23{,}641:4{,}7-4{,}6).$$

    Источник заимствования: Математика. 6 класс. Учебник для общеобразовательных организаций. В 2 частях. Часть 1 / – Мнемозина, 2019. – 107 c. ISBN 978-5-346-03720-0
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    Решение:

      • $$1)\ (7{,}061:2{,}3-2{,}2)\cdot(4{,}2+17{,}391:5{,}27)=(3{,}07-2{,}2)\cdot(4{,}2+17{,}391:5{,}27)=0{,}87\cdot(4{,}2+17{,}391:5{,}27)=0{,}87\cdot(4{,}2+3{,}3)=0{,}87\cdot7{,}5=6{,}525$$

      • $$\begin{array}{l}\begin{array}{}1)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}70{,}61\\69\\\hline\end{array}\begin{array}{|l}23\\\hline3{,}07\end{array}\\\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}16\\0\\\hline\end{array}\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}161\\161\\\hline\end{array}\end{array}\\\phantom{\ 0000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}2)\\[3.5ex]\\\end{array}\begin{array}{r}-\begin{array}{r}3{,}07\\2{,}20\\\hline\end{array}\\\begin{array}{r}0{,}87\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}3)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}1739{,}1\\1581\\\hline\end{array}\begin{array}{|l}527\\\hline3{,}3\end{array}\\\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}1581\\1581\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}4)\\[3.5ex]\\\end{array}\begin{array}{r}+\begin{array}{r}4{,}2\\3{,}3\\\hline\end{array}\\\begin{array}{r}7{,}5\end{array}\end{array}$$
      • $$\begin{array}{}5)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}0{,}87\\7{,}5\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}435\\609\phantom{0}\\\hline\end{array}\\\begin{array}{r}6{,}525\end{array}\end{array}\end{array}$$

      • $$2)\ (3{,}7+14{,}058:6{,}39)\cdot(23{,}641:4{,}7-4{,}6)=(3{,}7+2{,}2)\cdot(23{,}641:4{,}7-4{,}6)=5{,}9\cdot(23{,}641:4{,}7-4{,}6)=5{,}9\cdot(5{,}03-4{,}6)=5{,}9\cdot0{,}43=2{,}537$$

      • $$\begin{array}{l}\begin{array}{}1)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}1405{,}8\\1278\\\hline\end{array}\begin{array}{|l}639\\\hline2{,}2\end{array}\\\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}1278\\1278\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}2)\\[3.5ex]\\\end{array}\begin{array}{r}+\begin{array}{r}3{,}7\\2{,}2\\\hline\end{array}\\\begin{array}{r}5{,}9\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}3)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}236{,}41\\235\\\hline\end{array}\begin{array}{|l}47\\\hline5{,}03\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}14\\0\\\hline\end{array}\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}141\\141\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}4)\\[3.5ex]\\\end{array}\begin{array}{r}-\begin{array}{r}5{,}03\\4{,}60\\\hline\end{array}\\\begin{array}{r}0{,}43\end{array}\end{array}$$
      • $$\begin{array}{}5)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}0{,}43\\5{,}9\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}387\\215\phantom{0}\\\hline\end{array}\\\begin{array}{r}2{,}537\end{array}\end{array}\end{array}$$