Задание 146

Вычислите:

$${\largeа)}\ 3{,}(27)\cdot5-3{,}(27)\cdot4;$$
$${\largeб)}\ 5{,}(21)\cdot7+5{,}(21)\cdot3;$$

$${\largeв)}\ 3{,}(5)\cdot7{,}3-7{,}3\cdot3{,}(5);$$
$${\largeг)}\ 2{,}(7)\cdot5{,}41-5{,}41\cdot2{,}(7);$$

$${\largeд)}\ 13{,}(13)-13{,}(13);$$
$${\largeе)}\ {-}1\cdot3{,}(51);$$

$${\largeж)}\ 0\cdot5{,}1234567{...};$$
$${\largeз)}\ 1\cdot({-}5{,}1234567{...});$$

$${\largeи)}\ 1\cdot\frac{17}{19};$$
$${\largeк)}\ 3\cdot\frac{1}{3};$$

$${\largeл)}\ {-}3{,}4\cdot\frac{1}{{-}3{,}4};$$
$${\largeм)}\ {-}5\cdot\frac{1}{8};$$

$${\largeн)}\ 11{,}101101110{...}+({-}11{,}101101110{...}).$$

Источник заимствования: Алгебра. 7 класс. Учебник для общеобразовательных организаций / С.М. Никольский, М.К. Потапов, Н.Н. Решетников, А.В. Шевкин – Просвещение, 2013. – 37 c. ISBN 978-5-09-027739-6

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Решение:

$${\largeа)}\ 3{,}(27)\cdot5-3{,}(27)\cdot4=3{,}(27)\cdot(5-4)=3{,}(27)\cdot1=3{,}(27)$$

$${\largeб)}\ 5{,}(21)\cdot7+5{,}(21)\cdot3=5{,}(21)\cdot(7+3)=5{,}(21)\cdot10=52{,}(21)$$

$${\largeв)}\ 3{,}(5)\cdot7{,}3-7{,}3\cdot3{,}(5)=3{,}(5)\cdot(7{,}3-7{,}3)=3{,}(5)\cdot0=0$$

$${\largeг)}\ 2{,}(7)\cdot5{,}41-5{,}41\cdot2{,}(7)=2{,}(7)\cdot(5{,}41-5{,}41)=2{,}(7)\cdot0=0$$

$${\largeд)}\ 13{,}(13)-13{,}(13)=0$$

$${\largeе)}\ {-}1\cdot3{,}(51)={-}3{,}(51)$$

$${\largeж)}\ 0\cdot5{,}1234567{...}=0$$

$${\largeз)}\ 1\cdot({-}5{,}1234567{...})={-}5{,}1234567{...}$$

$${\largeи)}\ 1\cdot\frac{17}{19}=\frac{17}{19}$$

$${\largeк)}\ 3\cdot\frac{1}{3}=1$$

$${\largeл)}\ {-}3{,}4\cdot\frac{1}{{-}3{,}4}=1$$

$${\largeм)}\ {-}5\cdot\frac{1}{8}={-}\frac{5}{8}$$

$${\largeн)}\ 11{,}101101110{...}+({-}11{,}101101110{...})=0$$

Задание 145Задание 147