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МатематикаРешебник по алгебре 7 класс Мерзляк ФГОС Задание 127
Задание 127
\[\boxed{\text{127\ (127).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[- 3 \cdot ( - 4) + 2 = 12 + 2 = 14.\]
\[- 3 \cdot ( - 3) + 2 = 9 + 2 = 11.\]
\[- 3 \cdot ( - 2) + 2 = 6 + 2 = 8.\]
\[- 3 \cdot ( - 1) + 2 = \ 3 + 2 = 5.\]
\[- 3 \cdot 0 + 2 = 0 + 2 = 2.\]
\[- 3 \cdot 1 + 2 = - 3 + 2 = - 1.\]
\[- 3 \cdot 2 + 2 = - 6 + 2 = - 4.\]
\[- 3 \cdot 3 + 2 = - 9 + 2 = - 7.\]
\[- 3 \cdot 4 + 2 = - 12 + 2 = - 10.\]
| \[x\] | \[- 4\] | \[- 3\] | \[- 2\] | \[- 1\] | \[0\] | \[1\] | \[2\] | \[3\] | \[4\] |
|---|---|---|---|---|---|---|---|---|---|
| \[- 3x + 2\] | \[14\] | \[11\] | \[8\] | \[5\] | \[2\] | \[- 1\] | \[- 4\] | \[- 7\] | \[- 10\] |
\[\boxed{\text{127.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ 3ax = 12 - x;\ \ \ \ x = - 9\]
\[3 \cdot ( - 9) \cdot a = 12 - ( - 9)\]
\[- 27a = 12 + 9\]
\[- 27a = 21\]
\[a = - \frac{21}{27}\]
\[a = - \frac{7}{9}\]
\[Ответ:при\ a = - \frac{7}{9}.\]
\[2)\ (5a + 2) \cdot x = 8 - 2a;\ \ \ \ \]
\[\ x = 2\]
\[2 \cdot (5a + 2) = 8 - 2a\]
\[10a + 4 = 8 - 2a\]
\[10a + 2a = 8 - 4\]
\[12a = 4\]
\[a = \frac{4}{12}\]
\[a = \frac{1}{3}\]
\[Ответ:при\ a = \frac{1}{3}.\]
