Номер 179, страница 55 - гдз по алгебре 9 класс учебник Солтан, Солтан

Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019

Авторы: Солтан Г. Н., Солтан А. Е., Жумадилова А. Ж.

Тип: Учебник

Издательство: Кокшетау

Год издания: 2019 - 2026

ISBN: 978-601-317-424-2

Рекомендовано Министерством образования и науки Республики Казахстан

I. Уравнения, неравенства с двумя переменными и их системы. 6. Упражнения на повторение раздела «Уравнения, неравенства с двумя переменными и их системы» - номер 179, страница 55.

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            "text" => "<p><strong>а)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} x + y \le 2 \\ x + 2 \ge 0 \end{cases} $</p><p>Преобразуем каждое неравенство, чтобы выразить $y$ или $x$:</p><p><strong>1.</strong> Из первого неравенства $x + y \le 2$ получаем $y \le -x + 2$. Это неравенство описывает все точки на координатной плоскости, которые лежат на прямой $y = -x + 2$ и ниже нее. Граничная прямая $y = -x + 2$ строится по двум точкам, например, $(0, 2)$ и $(2, 0)$. Так как неравенство нестрогое ($\le$), прямая является частью решения и изображается сплошной линией.</p><p><strong>2.</strong> Из второго неравенства $x + 2 \ge 0$ получаем $x \ge -2$. Это неравенство описывает все точки, которые лежат на вертикальной прямой $x = -2$ и правее нее. Так как неравенство нестрогое ($\ge$), эта прямая также является частью решения и изображается сплошной линией.</p><p>Решением системы является пересечение (общая область) этих двух полуплоскостей. Чтобы найти вершину полученной угловой области, найдем точку пересечения граничных прямых $y = -x + 2$ и $x = -2$. Подставив $x = -2$ в уравнение первой прямой, получим $y = -(-2) + 2 = 4$. Таким образом, вершина находится в точке $(-2, 4)$.</p><p><strong>Ответ:</strong> Фигура, являющаяся решением системы, — это неограниченная область на плоскости (угол), ограниченная слева прямой $x = -2$ и сверху прямой $y = -x + 2$. Границы области (сами прямые) включены в решение. Вершина угла находится в точке $(-2, 4)$.</p><p><strong>б)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} y - 3x - 4 \le 0 \\ 2y - 6x + 8 \ge 0 \end{cases} $</p><p>Преобразуем оба неравенства:</p><p><strong>1.</strong> $y - 3x - 4 \le 0 \implies y \le 3x + 4$. Это полуплоскость, лежащая на прямой $y = 3x + 4$ и ниже нее.</p><p><strong>2.</strong> $2y - 6x + 8 \ge 0$. Разделим все члены на 2: $y - 3x + 4 \ge 0 \implies y \ge 3x - 4$. Это полуплоскость, лежащая на прямой $y = 3x - 4$ и выше нее.</p><p>Граничные прямые $y = 3x + 4$ и $y = 3x - 4$ имеют одинаковый угловой коэффициент $k=3$, что означает, что они параллельны.</p><p>Решением системы является пересечение двух полуплоскостей, то есть область, заключенная между этими двумя параллельными прямыми. Поскольку оба неравенства нестрогие ($\le$ и $\ge$), сами граничные прямые включаются в решение.</p><p><strong>Ответ:</strong> Фигура представляет собой бесконечную полосу, заключенную между параллельными прямыми $y = 3x + 4$ и $y = 3x - 4$. Границы полосы включены в решение.</p><p><strong>в)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} y &lt; \frac{4}{x} \\ y \ge x^2 \end{cases} $</p><p><strong>1.</strong> Первое неравенство $y &lt; \frac{4}{x}$ задает область под графиком функции $y = \frac{4}{x}$ (гипербола). Так как неравенство строгое ($<$), точки на самой гиперболе не являются частью решения, и она изображается пунктирной линией. Отметим, что $x \ne 0$.</p><p><strong>2.</strong> Второе неравенство $y \ge x^2$ задает область на параболе $y = x^2$ и выше нее. Так как неравенство нестрогое ($\ge$), сама парабола включена в решение и изображается сплошной линией.</p><p>Решением системы являются точки $(x, y)$, которые удовлетворяют обоим условиям, то есть $x^2 \le y &lt; \frac{4}{x}$. Это возможно только в случае, если $x^2 &lt; \frac{4}{x}$.</p><p>- При $x &gt; 0$ умножаем на $x$, получаем $x^3 &lt; 4$, откуда $0 &lt; x &lt; \sqrt[3]{4}$.</p><p>- При $x &lt; 0$ умножаем на $x$ и меняем знак неравенства: $x^3 &gt; 4$. Это неравенство не имеет решений для отрицательных $x$, так как куб отрицательного числа всегда отрицателен.</p><p>Таким образом, искомая фигура существует только в первой координатной четверти, где $0 &lt; x &lt; \sqrt[3]{4}$. Она ограничена снизу параболой $y=x^2$ и сверху гиперболой $y=4/x$. Найдем точку пересечения границ: $x^2 = \frac{4}{x} \implies x^3 = 4 \implies x = \sqrt[3]{4}$. Координата $y$ в этой точке: $y = (\sqrt[3]{4})^2 = \sqrt[3]{16}$.</p><p><strong>Ответ:</strong> Фигура является конечной областью в первой координатной четверти. Она ограничена снизу дугой параболы $y=x^2$ (включительно), сверху — дугой гиперболы $y=4/x$ (не включительно). Слева область неявно ограничена осью $Oy$ (не включительно). Область существует в интервале $0 &lt; x \le \sqrt[3]{4}$.</p><p><strong>г)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} |x| &lt; 2 \\ x^2 + y^2 &lt; 9 \end{cases} $</p><p><strong>1.</strong> Неравенство $|x| &lt; 2$ эквивалентно $-2 &lt; x &lt; 2$. Оно задает открытую вертикальную полосу между прямыми $x = -2$ и $x = 2$. Сами прямые не входят в решение.</p><p><strong>2.</strong> Неравенство $x^2 + y^2 &lt; 9$ (или $x^2 + y^2 &lt; 3^2$) задает множество точек внутри круга с центром в начале координат $(0, 0)$ и радиусом 3. Сама окружность не входит в решение.</p><p>Решением системы является пересечение этих двух областей — это центральная часть круга, вырезанная вертикальной полосой. Поскольку все неравенства строгие, границы полученной фигуры не включаются в решение.</p><p>Найдем точки пересечения граничных линий $x^2+y^2=9$ с прямыми $x=\pm 2$:</p><p>При $x = \pm 2$, имеем $(\pm 2)^2 + y^2 = 9 \implies 4 + y^2 = 9 \implies y^2 = 5 \implies y = \pm\sqrt{5}$.</p><p>Фигура ограничена двумя вертикальными отрезками (на прямых $x=-2$ и $x=2$ между $y=-\sqrt{5}$ и $y=\sqrt{5}$) и двумя дугами окружности, соединяющими концы этих отрезков.</p><p><strong>Ответ:</strong> Фигура представляет собой открытый сегмент круга: внутренность круга с центром в $(0,0)$ и радиусом 3, находящаяся в полосе $-2 &lt; x &lt; 2$. Границы фигуры (дуги окружности и вертикальные отрезки) не включены в решение.</p>"
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            "text" => "<p><strong>а)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} x + y \le 2 \\ x + 2 \ge 0 \end{cases} $</p><p>Преобразуем каждое неравенство, чтобы выразить $y$ или $x$:</p><p><strong>1.</strong> Из первого неравенства $x + y \le 2$ получаем $y \le -x + 2$. Это неравенство описывает все точки на координатной плоскости, которые лежат на прямой $y = -x + 2$ и ниже нее. Граничная прямая $y = -x + 2$ строится по двум точкам, например, $(0, 2)$ и $(2, 0)$. Так как неравенство нестрогое ($\le$), прямая является частью решения и изображается сплошной линией.</p><p><strong>2.</strong> Из второго неравенства $x + 2 \ge 0$ получаем $x \ge -2$. Это неравенство описывает все точки, которые лежат на вертикальной прямой $x = -2$ и правее нее. Так как неравенство нестрогое ($\ge$), эта прямая также является частью решения и изображается сплошной линией.</p><p>Решением системы является пересечение (общая область) этих двух полуплоскостей. Чтобы найти вершину полученной угловой области, найдем точку пересечения граничных прямых $y = -x + 2$ и $x = -2$. Подставив $x = -2$ в уравнение первой прямой, получим $y = -(-2) + 2 = 4$. Таким образом, вершина находится в точке $(-2, 4)$.</p><p><strong>Ответ:</strong> Фигура, являющаяся решением системы, — это неограниченная область на плоскости (угол), ограниченная слева прямой $x = -2$ и сверху прямой $y = -x + 2$. Границы области (сами прямые) включены в решение. Вершина угла находится в точке $(-2, 4)$.</p><p><strong>б)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} y - 3x - 4 \le 0 \\ 2y - 6x + 8 \ge 0 \end{cases} $</p><p>Преобразуем оба неравенства:</p><p><strong>1.</strong> $y - 3x - 4 \le 0 \implies y \le 3x + 4$. Это полуплоскость, лежащая на прямой $y = 3x + 4$ и ниже нее.</p><p><strong>2.</strong> $2y - 6x + 8 \ge 0$. Разделим все члены на 2: $y - 3x + 4 \ge 0 \implies y \ge 3x - 4$. Это полуплоскость, лежащая на прямой $y = 3x - 4$ и выше нее.</p><p>Граничные прямые $y = 3x + 4$ и $y = 3x - 4$ имеют одинаковый угловой коэффициент $k=3$, что означает, что они параллельны.</p><p>Решением системы является пересечение двух полуплоскостей, то есть область, заключенная между этими двумя параллельными прямыми. Поскольку оба неравенства нестрогие ($\le$ и $\ge$), сами граничные прямые включаются в решение.</p><p><strong>Ответ:</strong> Фигура представляет собой бесконечную полосу, заключенную между параллельными прямыми $y = 3x + 4$ и $y = 3x - 4$. Границы полосы включены в решение.</p><p><strong>в)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} y &lt; \frac{4}{x} \\ y \ge x^2 \end{cases} $</p><p><strong>1.</strong> Первое неравенство $y &lt; \frac{4}{x}$ задает область под графиком функции $y = \frac{4}{x}$ (гипербола). Так как неравенство строгое ($<$), точки на самой гиперболе не являются частью решения, и она изображается пунктирной линией. Отметим, что $x \ne 0$.</p><p><strong>2.</strong> Второе неравенство $y \ge x^2$ задает область на параболе $y = x^2$ и выше нее. Так как неравенство нестрогое ($\ge$), сама парабола включена в решение и изображается сплошной линией.</p><p>Решением системы являются точки $(x, y)$, которые удовлетворяют обоим условиям, то есть $x^2 \le y &lt; \frac{4}{x}$. Это возможно только в случае, если $x^2 &lt; \frac{4}{x}$.</p><p>- При $x &gt; 0$ умножаем на $x$, получаем $x^3 &lt; 4$, откуда $0 &lt; x &lt; \sqrt[3]{4}$.</p><p>- При $x &lt; 0$ умножаем на $x$ и меняем знак неравенства: $x^3 &gt; 4$. Это неравенство не имеет решений для отрицательных $x$, так как куб отрицательного числа всегда отрицателен.</p><p>Таким образом, искомая фигура существует только в первой координатной четверти, где $0 &lt; x &lt; \sqrt[3]{4}$. Она ограничена снизу параболой $y=x^2$ и сверху гиперболой $y=4/x$. Найдем точку пересечения границ: $x^2 = \frac{4}{x} \implies x^3 = 4 \implies x = \sqrt[3]{4}$. Координата $y$ в этой точке: $y = (\sqrt[3]{4})^2 = \sqrt[3]{16}$.</p><p><strong>Ответ:</strong> Фигура является конечной областью в первой координатной четверти. Она ограничена снизу дугой параболы $y=x^2$ (включительно), сверху — дугой гиперболы $y=4/x$ (не включительно). Слева область неявно ограничена осью $Oy$ (не включительно). Область существует в интервале $0 &lt; x \le \sqrt[3]{4}$.</p><p><strong>г)</strong></p><p>Рассмотрим систему неравенств:</p><p>$ \begin{cases} |x| &lt; 2 \\ x^2 + y^2 &lt; 9 \end{cases} $</p><p><strong>1.</strong> Неравенство $|x| &lt; 2$ эквивалентно $-2 &lt; x &lt; 2$. Оно задает открытую вертикальную полосу между прямыми $x = -2$ и $x = 2$. Сами прямые не входят в решение.</p><p><strong>2.</strong> Неравенство $x^2 + y^2 &lt; 9$ (или $x^2 + y^2 &lt; 3^2$) задает множество точек внутри круга с центром в начале координат $(0, 0)$ и радиусом 3. Сама окружность не входит в решение.</p><p>Решением системы является пересечение этих двух областей — это центральная часть круга, вырезанная вертикальной полосой. Поскольку все неравенства строгие, границы полученной фигуры не включаются в решение.</p><p>Найдем точки пересечения граничных линий $x^2+y^2=9$ с прямыми $x=\pm 2$:</p><p>При $x = \pm 2$, имеем $(\pm 2)^2 + y^2 = 9 \implies 4 + y^2 = 9 \implies y^2 = 5 \implies y = \pm\sqrt{5}$.</p><p>Фигура ограничена двумя вертикальными отрезками (на прямых $x=-2$ и $x=2$ между $y=-\sqrt{5}$ и $y=\sqrt{5}$) и двумя дугами окружности, соединяющими концы этих отрезков.</p><p><strong>Ответ:</strong> Фигура представляет собой открытый сегмент круга: внутренность круга с центром в $(0,0)$ и радиусом 3, находящаяся в полосе $-2 &lt; x &lt; 2$. Границы фигуры (дуги окружности и вертикальные отрезки) не включены в решение.</p>"
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№179 (с. 55)
Условие. №179 (с. 55)
скриншот условия
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 55, номер 179, Условие

179. Постройте в координатной плоскости фигуру, координаты каждой точки которой являются решениями системы неравенств:

a) $\begin{cases} x + y \le 2, \\ x + 2 \ge 0; \end{cases}$

б) $\begin{cases} y - 3x - 4 \le 0, \\ 2y - 6x + 8 \ge 0; \end{cases}$

в) $\begin{cases} y < \frac{4}{x}, \\ y \ge x^2; \end{cases}$

г) $\begin{cases} |x| < 2, \\ x^2 + y^2 < 9. \end{cases}$

Решение. №179 (с. 55)
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 55, номер 179, Решение
Решение 2 (rus). №179 (с. 55)

а)

Рассмотрим систему неравенств:

$ \begin{cases} x + y \le 2 \\ x + 2 \ge 0 \end{cases} $

Преобразуем каждое неравенство, чтобы выразить $y$ или $x$:

1. Из первого неравенства $x + y \le 2$ получаем $y \le -x + 2$. Это неравенство описывает все точки на координатной плоскости, которые лежат на прямой $y = -x + 2$ и ниже нее. Граничная прямая $y = -x + 2$ строится по двум точкам, например, $(0, 2)$ и $(2, 0)$. Так как неравенство нестрогое ($\le$), прямая является частью решения и изображается сплошной линией.

2. Из второго неравенства $x + 2 \ge 0$ получаем $x \ge -2$. Это неравенство описывает все точки, которые лежат на вертикальной прямой $x = -2$ и правее нее. Так как неравенство нестрогое ($\ge$), эта прямая также является частью решения и изображается сплошной линией.

Решением системы является пересечение (общая область) этих двух полуплоскостей. Чтобы найти вершину полученной угловой области, найдем точку пересечения граничных прямых $y = -x + 2$ и $x = -2$. Подставив $x = -2$ в уравнение первой прямой, получим $y = -(-2) + 2 = 4$. Таким образом, вершина находится в точке $(-2, 4)$.

Ответ: Фигура, являющаяся решением системы, — это неограниченная область на плоскости (угол), ограниченная слева прямой $x = -2$ и сверху прямой $y = -x + 2$. Границы области (сами прямые) включены в решение. Вершина угла находится в точке $(-2, 4)$.

б)

Рассмотрим систему неравенств:

$ \begin{cases} y - 3x - 4 \le 0 \\ 2y - 6x + 8 \ge 0 \end{cases} $

Преобразуем оба неравенства:

1. $y - 3x - 4 \le 0 \implies y \le 3x + 4$. Это полуплоскость, лежащая на прямой $y = 3x + 4$ и ниже нее.

2. $2y - 6x + 8 \ge 0$. Разделим все члены на 2: $y - 3x + 4 \ge 0 \implies y \ge 3x - 4$. Это полуплоскость, лежащая на прямой $y = 3x - 4$ и выше нее.

Граничные прямые $y = 3x + 4$ и $y = 3x - 4$ имеют одинаковый угловой коэффициент $k=3$, что означает, что они параллельны.

Решением системы является пересечение двух полуплоскостей, то есть область, заключенная между этими двумя параллельными прямыми. Поскольку оба неравенства нестрогие ($\le$ и $\ge$), сами граничные прямые включаются в решение.

Ответ: Фигура представляет собой бесконечную полосу, заключенную между параллельными прямыми $y = 3x + 4$ и $y = 3x - 4$. Границы полосы включены в решение.

в)

Рассмотрим систему неравенств:

$ \begin{cases} y < \frac{4}{x} \\ y \ge x^2 \end{cases} $

1. Первое неравенство $y < \frac{4}{x}$ задает область под графиком функции $y = \frac{4}{x}$ (гипербола). Так как неравенство строгое ($<$), точки на самой гиперболе не являются частью решения, и она изображается пунктирной линией. Отметим, что $x \ne 0$.

2. Второе неравенство $y \ge x^2$ задает область на параболе $y = x^2$ и выше нее. Так как неравенство нестрогое ($\ge$), сама парабола включена в решение и изображается сплошной линией.

Решением системы являются точки $(x, y)$, которые удовлетворяют обоим условиям, то есть $x^2 \le y < \frac{4}{x}$. Это возможно только в случае, если $x^2 < \frac{4}{x}$.

- При $x > 0$ умножаем на $x$, получаем $x^3 < 4$, откуда $0 < x < \sqrt[3]{4}$.

- При $x < 0$ умножаем на $x$ и меняем знак неравенства: $x^3 > 4$. Это неравенство не имеет решений для отрицательных $x$, так как куб отрицательного числа всегда отрицателен.

Таким образом, искомая фигура существует только в первой координатной четверти, где $0 < x < \sqrt[3]{4}$. Она ограничена снизу параболой $y=x^2$ и сверху гиперболой $y=4/x$. Найдем точку пересечения границ: $x^2 = \frac{4}{x} \implies x^3 = 4 \implies x = \sqrt[3]{4}$. Координата $y$ в этой точке: $y = (\sqrt[3]{4})^2 = \sqrt[3]{16}$.

Ответ: Фигура является конечной областью в первой координатной четверти. Она ограничена снизу дугой параболы $y=x^2$ (включительно), сверху — дугой гиперболы $y=4/x$ (не включительно). Слева область неявно ограничена осью $Oy$ (не включительно). Область существует в интервале $0 < x \le \sqrt[3]{4}$.

г)

Рассмотрим систему неравенств:

$ \begin{cases} |x| < 2 \\ x^2 + y^2 < 9 \end{cases} $

1. Неравенство $|x| < 2$ эквивалентно $-2 < x < 2$. Оно задает открытую вертикальную полосу между прямыми $x = -2$ и $x = 2$. Сами прямые не входят в решение.

2. Неравенство $x^2 + y^2 < 9$ (или $x^2 + y^2 < 3^2$) задает множество точек внутри круга с центром в начале координат $(0, 0)$ и радиусом 3. Сама окружность не входит в решение.

Решением системы является пересечение этих двух областей — это центральная часть круга, вырезанная вертикальной полосой. Поскольку все неравенства строгие, границы полученной фигуры не включаются в решение.

Найдем точки пересечения граничных линий $x^2+y^2=9$ с прямыми $x=\pm 2$:

При $x = \pm 2$, имеем $(\pm 2)^2 + y^2 = 9 \implies 4 + y^2 = 9 \implies y^2 = 5 \implies y = \pm\sqrt{5}$.

Фигура ограничена двумя вертикальными отрезками (на прямых $x=-2$ и $x=2$ между $y=-\sqrt{5}$ и $y=\sqrt{5}$) и двумя дугами окружности, соединяющими концы этих отрезков.

Ответ: Фигура представляет собой открытый сегмент круга: внутренность круга с центром в $(0,0)$ и радиусом 3, находящаяся в полосе $-2 < x < 2$. Границы фигуры (дуги окружности и вертикальные отрезки) не включены в решение.

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