§ 4. Упражнение 897. «Математика. 6 класс. Учебник для общеобразовательных организаций. В 2 частях. Часть 1»

    Упражнение 897

    Найдите значение выражения:

      • $${\largeа)}\ 150{,}88:(3{,}2\cdot2{,}3)-60{,}27:(4{,}1\cdot1{,}4);$$$${\largeб)}\ 592{,}92:(2{,}7\cdot7{,}2)-102{,}48:(6{,}1\cdot1{,}6).$$

    Источник заимствования: Математика. 6 класс. Учебник для общеобразовательных организаций. В 2 частях. Часть 1 / – Мнемозина, 2019. – 159 c. ISBN 978-5-346-03720-0
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    Решение:

      • $${\largeа)}\ 150{,}88:(3{,}2\cdot2{,}3)-60{,}27:(4{,}1\cdot1{,}4)=150{,}88:7{,}36-60{,}27:(4{,}1\cdot1{,}4)=150{,}88:7{,}36-60{,}27:5{,}74=20{,}5-60{,}27:5{,}74=20{,}5-10{,}5=10$$

      • $$\begin{array}{}1)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}3{,}2\\2{,}3\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}96\\64\phantom{0}\\\hline\end{array}\\\begin{array}{r}7{,}36\end{array}\end{array}\end{array}$$
      • $$\begin{array}{}2)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}4{,}1\\1{,}4\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}164\\41\phantom{0}\\\hline\end{array}\\\begin{array}{r}5{,}74\end{array}\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}3)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}15088{,}0\\1472\\\hline\end{array}\begin{array}{|l}736\\\hline20{,}5\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}368\\0\\\hline\end{array}\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}3680\\3680\\\hline\end{array}\end{array}\\\phantom{\ 000000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}4)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}6027{,}0\\574\\\hline\end{array}\begin{array}{|l}574\\\hline10{,}5\end{array}\\\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}287\\0\\\hline\end{array}\end{array}\\\phantom{\ 000}\begin{array}{r}-\begin{array}{r}2870\\2870\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}5)\\[3.5ex]\\\end{array}\begin{array}{r}-\begin{array}{r}20{,}5\\10{,}5\\\hline\end{array}\\\begin{array}{r}10{,}0\end{array}\end{array}$$

      • $${\largeб)}\ 592{,}92:(2{,}7\cdot7{,}2)-102{,}48:(6{,}1\cdot1{,}6)=592{,}92:19{,}44-102{,}48:(6{,}1\cdot1{,}6)=592{,}92:19{,}44-102{,}48:9{,}76=30{,}5-102{,}48:9{,}76=30{,}5-10{,}5=20$$

      • $$\begin{array}{}1)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}2{,}7\\7{,}2\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}54\\189\phantom{0}\\\hline\end{array}\\\begin{array}{r}19{,}44\end{array}\end{array}\end{array}$$
      • $$\begin{array}{}2)\\[10.5ex]\\\end{array}\begin{array}{r}\times\begin{array}{r}6{,}1\\1{,}6\\\hline\end{array}\phantom{0}\\\begin{array}{r}+\begin{array}{r}366\\61\phantom{0}\\\hline\end{array}\\\begin{array}{r}9{,}76\end{array}\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}3)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}59292{,}0\\5832\\\hline\end{array}\begin{array}{|l}1944\\\hline30{,}5\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}972\\0\\\hline\end{array}\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}9720\\9720\\\hline\end{array}\end{array}\\\phantom{\ 000000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{l}\begin{array}{}4)\\[0.2ex]\\\end{array}\begin{array}{r}-\begin{array}{l}10248{,}0\\\phantom{0}976\\\hline\end{array}\begin{array}{|l}976\\\hline10{,}5\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}488\\0\\\hline\end{array}\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}4880\\4880\\\hline\end{array}\end{array}\\\phantom{\ 000000000}\begin{array}{r}0\end{array}\end{array}$$
      • $$\begin{array}{}5)\\[3.5ex]\\\end{array}\begin{array}{r}-\begin{array}{r}30{,}5\\10{,}5\\\hline\end{array}\\\begin{array}{r}20{,}0\end{array}\end{array}$$