Решебник по алгебре 8 класс Макарычев ФГОС Задание 1279

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\[\frac{\overline{\text{ab}}}{a + b} = z,\ \ b > a\]

\[\frac{10a + b}{a + b} = z\]

\[10a + b = z(a + b)\]

\[10a + b = za + zb\]

\[10a - za = zb - b\]

\[a(10 - z) = b(z - 1)\]

\[так\ как\ b > a:\]

\[10 - z > z - 1\]

\[- 2z > - 11\]

\[z < 5,5\]

\[z \in \left\{ 1,\ 2,\ 3,\ 4,\ 5 \right\}\]

\[z = 1 \Longrightarrow 10a + b = a + b,\]

\[\ \ 9a = 0,\ \ \varnothing\]

\[z = 2 \Longrightarrow a(10 - 2) = b,\]

\[\ \ a = 1,\ \ b = 8\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8a = b\]

\[z = 3 \Longrightarrow a(10 - 3) = b(3 - 1),\]

\[\ \ a = 2,\ \ b = 7\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7a = 2b\]

\[z = 4 \Longrightarrow a(10 - 4) = (4 - 1)b,\]

\[\ \ a = 1,\ \ b = 2\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6a = 3b,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ a = 2,\ \ b = 4\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2a = b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ a = 3,\ \ b = 6\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \ \ \ \ \ \ \ \ \ \ a = 4,\ \ b = 8\]

\[z = 5 \Longrightarrow 5a = 4b,\ \ a = 4,\]

\[\ \ b = 5\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{5a}{4} = b\]

\[\overline{\text{ab}} = \left\{ 12,\ 18,\ 24,\ 27,\ 36,\ 45,\ 48 \right\}.\]