ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 948

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\[\textbf{а)}\ \frac{2x}{5} > 1\]

\[2x > 5\]

\[x > 2,5\]

\[x \in (2,5;\ + \infty)\]

\[\textbf{б)}\ \frac{x}{3} < 2\]

\[x < 6\]

\[x \in ( - \infty;6)\]

\[\textbf{в)}\ \frac{6x}{7} \geq 0\]

\[6x \geq 0\]

\[x \geq 0\]

\[x \in \lbrack 0;\ + \infty)\]

\[\textbf{г)}\ \frac{3x - 1}{4} > 2\]

\[3x - 1 > 8\]

\[3x > 9\]

\[x > 3\]

\[x \in (3;\ + \infty)\]

\[\textbf{д)}\ 2 > \frac{6 - x}{5}\]

\[10 > 6 - x\]

\[x > - 10 + 6\]

\[x > 6 - 10\]

\[x > - 4\]

\[x \in ( - 4;\ + \infty)\]

\[\textbf{е)}\ \frac{2 + 3x}{18} < 0\]

\[2 + 3x < 0\]

\[3x < - 2\]

\[x < - \frac{2}{3}\]

\[x \in \left( - \infty;\ - \frac{2}{3} \right)\]

\[\textbf{ж)}\ \frac{12 - 7x}{42} \geq 0\]

\[12 - 7x \geq 0\]

\[- 7x \geq - 12\]

\[x \leq \frac{12}{7}\]

\[x \in \left( - \infty;1\frac{5}{7} \right\rbrack\]

\[\textbf{з)}\frac{1}{3} \cdot (x + 15) > 4\]

\[x + 15 > 12\]

\[x > 12 - 15\]

\[x > - 3\]

\[x \in ( - 3;\ + \infty)\]

\[\textbf{и)}\ 6 \leq \frac{2}{7} \cdot (x + 4)\]

\[21 \leq x + 4\]

\[21 - 4 \leq x\]

\[x \geq 17\]

\[x \in \lbrack 17;\ + \infty)\]