Страница 229 номер 789, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

Задачи для внеклассной работы. Страница 229. Номер 789

Задание / условие:

Решить систему уравнений: 1) \(\begin{cases}|x - 1| + |y - 5| = 1,\\ y = 5 + |x - 1|;\end{cases}\) 2) \(\begin{cases}\sqrt{5y - x} + x = 3,\\ \sqrt{2y - x} + x + y = 3;\end{cases}\) 3) \(\begin{cases}\sqrt{y + 7x} + \sqrt{y + 2x} = 5,\\ \sqrt{y + 2x} - y + x = 1;\end{cases}\) 4) \(\begin{cases}\sqrt{25 - x^2} - \sqrt{25 - y^2} = \sqrt{8},\\ \sqrt{25 - x^2} + \sqrt{25 - y^2} = \sqrt{16 + (x + y)^2}.\end{cases}\)

Решение:

1) \(\begin{cases}|x - 1| + |y - 5| = 1 \\ y = 5 + |x - 1|\end{cases}\)
\(\begin{cases}|x - 1| = 1 - |y - 5| \\ y - 5 + |y - 5| = 1\end{cases},\)

Второе уравнение при \(y < 5\):

\(y - 5 + 5 - y = 1\) – не имеет решений;

при \(y \geqslant 5\):

\(y - 5 + y - 5 = 1,\)
\(2y = 11,\)
\(y = 5{,}5;\)
\(\begin{cases}y = 5{,}5 \\ |x - 1| = 1 - |5{,}5 - 5|\end{cases},\)
\(\begin{cases}y = 5{,}5 \\ x = 1{,}5\end{cases}\) или \(\begin{cases}y = 5{,}5 \\ x = 0{,}5\end{cases}\).

Ответ: \((1{,}5; 5{,}5)\), \((0{,}5; 5{,}5)\);

2) \(\begin{cases}\sqrt{5y - x} + x = 3 \\ \sqrt{2y - x} + x + y = 3\end{cases}\)
\(\sqrt{2y - x} + x + y - \sqrt{5y - x} - x = 3 - 3,\)
\(\sqrt{2y - x} + y = \sqrt{5y - x},\)
\(2y - x + 2y\sqrt{2y - x} + y^{2} = 5y - x,\)
\(3y - 2y\sqrt{2y - x} - y^{2} = 0,\)
\(y\left(3 - 2\sqrt{2y - x} - y\right) = 0,\)
\(y_{1} = 0,\)
\(2\sqrt{2y - x} = 3 - y,\)
\(4(2y - x) = 9 - 6y + y^{2},\)
\(8y - 4x - 9 + 6y - y^{2} = 0,\)
\(4x = 8y - 9 + 6y - y^{2},\)
\(x = \dfrac{14y - 9 - y^{2}}{4};\)
\(\sqrt{5y - \dfrac{14y - 9 - y^{2}}{4}} + \dfrac{14y - 9 - y^{2}}{4} = 3,\)
\(\sqrt{\dfrac{y^{2} + 6y + 9}{4}} + \dfrac{14y - 9 - y^{2}}{4} = 3,\)
\(y + 3 \geqslant 0\), т.к. \(\frac{y + 3}{2} = \frac{3 - y}{2} + y = \sqrt{2y - x} + y = \sqrt{5y - x} \geqslant 0\);
\(\dfrac{y + 3}{2} + \dfrac{14y - 9 - y^{2}}{4} - 3 = 0,\)
\(\dfrac{4y + 12 + 28y - 18 - 2y^{2} - 24}{8} = 0,\)
\(2y^{2} - 32y + 30 = 0,\)
\(y^{2} - 16y + 15 = 0,\)
\(y_{2{,}3} = \dfrac{16 \pm \sqrt{256 - 60}}{2},\)
\(y_{2} = 1, y_{3} = 15;\)
\(1.\enspace y = 0:\)
\(\sqrt{-x} + x = 3,\)
\(-x = 9 - 6x + x^{2},\)
\(x^{2} - 5x + 9 = 0,\)
\(D = 25 - 36 < 0\) – корней нет;
\(2.\enspace y = 1:\)
\(x = \dfrac{14y - 9 - y^{2}}{4} = \dfrac{14 - 9 - 1}{4} = 1,\)
\(\begin{cases}\sqrt{5 - 1} + 1 = 3 \\ \sqrt{2 - 1} + 1 + 1 = 3\end{cases}\) – верно;
\(3.\enspace y = 15:\)
\(x = \dfrac{14y - 9 - y^{2}}{4} = \dfrac{14 \cdot 15 - 9 - 225}{4} = -6;\)
\(\begin{cases}\sqrt{5 \cdot 15 + 6} - 6 = 3 \\ \sqrt{2 \cdot 15 + 6} - 6 + 15 = 3\end{cases}\) – не верно. Посторонний корень.

Ответ: \((1; 1)\);

3) \(\begin{cases}\sqrt{y + 7x} + \sqrt{y + 2x} = 5 \\ \sqrt{y + 2x} - y + x = 1\end{cases}\)
\(z = y - x, y = z + x\)
\(\begin{cases}\sqrt{z + x + 7x} + \sqrt{z + x + 2x} = 5 \\ \sqrt{z + x + 2x} - z - x + x = 1\end{cases},\)
\(\begin{cases}\sqrt{z + 8x} + \sqrt{z + 3x} = 5 \\ \sqrt{z + 3x} - z = 1\end{cases},\)
\(\begin{cases}\sqrt{z + 3x} = 1 + z \\ \sqrt{z + 8x} = 5 - 1 - z\end{cases},\)
\(\begin{cases}z + 3x = 1 + 2z + z^{2} \\ z + 8x = 16 - 8z + z^{2}\end{cases},\)
\(\begin{cases}x = \frac{1 + z + z^{2}}{3} \\ 16 - 9z + z^{2} = \frac{8\left(1 + z + z^{2}\right)}{3}\end{cases},\)
\(48 - 27z + 3z^{2} = 8 + 8z + 8z^{2},\)
\(5z^{2} + 35z - 40 = 0,\)
\(z^{2} + 7z - 8 = 0,\)
\(z_{1{,}2} = \dfrac{-7 \pm \sqrt{49 + 32}}{2},\)
\(z_{1} = 1, z_{2} = -8;\)
\(1.\enspace x = \dfrac{1 - 8 + 64}{3} = 19,\)
\(y = -8 + x = 11,\)
\(\begin{cases}\sqrt{11 + 7 \cdot 19} + \sqrt{11 + 2 \cdot 19} = 5 \\ \sqrt{11 + 2 \cdot 19} - 11 + 19 = 1\end{cases},\)
\(\begin{cases}\sqrt{144} + \sqrt{49} = 5 \\ \sqrt{49} - 11 + 19 = 1\end{cases}\) – не верно. Посторонний корень;
\(2.\enspace x = \dfrac{1 + 1 + 1}{3} = 1,\)
\(y = 1 + x = 1 + 1 = 2,\)
\(\begin{cases}\sqrt{2 + 7 \cdot 1} + \sqrt{2 + 2 \cdot 1} = 5 \\ \sqrt{2 + 2 \cdot 1} - 2 + 1 = 1\end{cases},\)
\(\begin{cases}\sqrt{9} + \sqrt{4} = 5 \\ \sqrt{4} - 1 = 1\end{cases}\) – верно.

Ответ: \((1; 2)\);

4) \(\begin{cases}\sqrt{25 - x^{2}} - \sqrt{25 - y^{2}} = \sqrt{8} \\ \sqrt{25 - x^{2}} + \sqrt{25 - y^{2}} = \sqrt{16 + (x + y)^{2}}\end{cases}\)
\(\begin{cases}25 - x^{2} - 2\sqrt{\left(25 - x^{2}\right)\left(25 - y^{2}\right)} + 25 - y^{2} = 8 \\ 25 - x^{2} + 2\sqrt{\left(25 - x^{2}\right)\left(25 - y^{2}\right)} + 25 - y^{2} = 16 + (x + y)^{2}\end{cases},\)

Сложим полученные уравнения:

\(100 - 2x^{2} - 2y^{2} = 24 + (x + y)^{2},\)
\(76 - 2\left(x^{2} + y^{2}\right) - (x + y)^{2} = 0;\)

Перемножим уравнения исходной системы:

\(25 - x^{2} + \sqrt{\left(25 - x^{2}\right)\left(25 - y^{2}\right)} -\)
\(- \sqrt{\left(25 - x^{2}\right)\left(25 - y^{2}\right)} - 25 + y^{2} =\)
\(= \sqrt{8\left(16 + (x + y)^{2}\right)},\)
\(y^{2} - x^{2} = \sqrt{8\left(16 + (x + y)^{2}\right)},\)

Возведём в квадрат при условии \(y^{2} - x^{2} \geqslant 0\):

\(\left((y - x)(y + x)\right)^{2} = 8\left(16 + (x + y)^{2}\right);\)

Из полученных уравнений составим систему:

\(\begin{cases}76 - 2\left(y^{2} + x^{2}\right) - (y + x)^{2} = 0 \\ (y - x)^{2}(y + x)^{2} = 128 + 8(y + x)^{2}\end{cases},\)
\((y + x)^{2} = u, (y - x)^{2} = v\)
\(2\left(y^{2} + x^{2}\right) + (y + x)^{2} = 2y^{2} + 2x^{2} + y^{2} + 2yx + x^{2} =\)
\(= 2y^{2} + 4yx + 2x^{2} + y^{2} - 2yx + x^{2} =\)
\(= 2(y + x)^{2} + (y - x)^{2} = 2u + v,\)
\(\begin{cases}76 = 2u + v \\ uv = 128 + 8u\end{cases},\)
\(\begin{cases}v = 76 - 2u \\ -2u^{2} + 76u - 128 - 8u = 0\end{cases},\)
\(u^{2} - 34u + 64 = 0,\)
\(u_{1{,}2} = \dfrac{34 \pm \sqrt{1156 - 256}}{2},\)
\(u_{1} = 2, u_{2} = 32,\)
\(v_{1} = 72, v_{2} = 12;\)
\(1.\enspace \begin{cases}(y + x)^{2} = 2 \\ (y - x)^{2} = 72\end{cases},\)
\(\begin{cases}y^{2} + 2yx + x^{2} = 2 \\ y^{2} - 2yx + x^{2} = 72\end{cases},\)
\(-4yx = 70,\)
\(yx = -\dfrac{35}{2},\)
\(x = -\dfrac{35}{2y},\)
\(\begin{cases}y^{2} - 35 + \frac{1225}{4y^{2}} = 2 \\ y^{2} + 35 + \frac{1225}{4y^{2}} = 72\end{cases},\)
\(\dfrac{4y^{4} - 140y^{2} + 1225 - 8y^{2}}{4y^{2}} = 0,\)
\(4y^{4} - 148y^{2} + 1225 = 0,\)
\(y_{1{,}2}^{2} = \dfrac{148 \pm \sqrt{21904 - 19600}}{8},\)
\(y_{1}^{2} = \dfrac{25}{2}, y_{2}^{2} = \dfrac{49}{2};\)
\(y = \pm \frac{5}{\sqrt{2}}\) или \(y = \pm \frac{7}{\sqrt{2}}\)

Тогда \(x = -\frac{35}{2y}\):

\(x = \mp \frac{7}{\sqrt{2}}\) или \(x = \mp \frac{5}{\sqrt{2}}\).
\(y^{2} - x^{2} \geqslant 0\) только при \(y = \pm \frac{7}{\sqrt{2}}\), \(x = \mp \frac{5}{\sqrt{2}}\).
\(2.\enspace \begin{cases}(y + x)^{2} = 32 \\ (y - x)^{2} = 12\end{cases},\)
\(\begin{cases}y^{2} + 2yx + x^{2} = 32 \\ y^{2} - 2yx + x^{2} = 12\end{cases},\)
\(4yx = 20,\)
\(x = \dfrac{5}{y},\)
\(y^{2} + 10 + \dfrac{25}{y^{2}} = 32,\)
\(y^{4} + 10y^{2} + 25 - 32y^{2} = 0,\)
\(y^{4} - 22y^{2} + 25 = 0,\)
\(y_{1{,}2}^{2} = \dfrac{22 \pm \sqrt{484 - 100}}{2} = 11 \pm 4\sqrt{6};\)
\(y_{1{,}2} = \pm \sqrt{11 \pm 4\sqrt{6}},\)
\(x_{1{,}2} = \pm \dfrac{5}{\sqrt{11 \pm 4\sqrt{6}}}.\)
\(x^{2} = \frac{25}{11 \pm 4\sqrt{6}} = \frac{25\left(11 \mp 4\sqrt{6}\right)}{121 - 96} = 11 \mp 4\sqrt{6}\);
\(y^{2} - x^{2} \geqslant 0\) только при \(y = \pm \sqrt{11 + 4\sqrt{6}}\), \(x = \pm \frac{5}{\sqrt{11 + 4\sqrt{6}}}\).