Страница 228 номер 787, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(\begin{cases}x^{3} - y^{3} = 61(x - y) \\ (x + 1)(y + 1) = 12\end{cases}\)
\(\begin{cases}(x - y)\left(x^{2} + xy + y^{2}\right) = 61(x - y) \\ (x + 1)(y + 1) = 12\end{cases},\)
\(\begin{cases}x - y = 0 \\ (x + 1)(y + 1) = 12\end{cases}\) или \(\begin{cases}x^{2} + xy + y^{2} = 61 \\ (x + 1)(y + 1) = 12\end{cases}\)
\(1.\enspace \begin{cases}x = y \\ (y + 1)^{2} = 12\end{cases}\)
\(y^{2} + 2y - 11 = 0,\)
\(y_{1{,}2} = \dfrac{-2 \pm \sqrt{4 + 44}}{2} = \dfrac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3},\)
\(\begin{cases}y_{1} = 2\sqrt{3} - 1 \\ x_{1} = 2\sqrt{3} - 1\end{cases},\)
\(\begin{cases}y_{2} = -2\sqrt{3} - 1 \\ x_{2} = -2\sqrt{3} - 1\end{cases};\)
\(2.\enspace \begin{cases}x^{2} + xy + y^{2} + xy + x + y + 1 = 73 \\ xy + x + y + 1 = 12\end{cases},\)
\(\begin{cases}(x + y)^{2} + x + y = 72 \\ xy + x + y = 11\end{cases},\)
\(x + y = z,\)
\(z^{2} + z - 72 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 288}}{2},\)
\(z_{1} = 8, z_{2} = -9;\)
\(\begin{cases}x + y = 8 \\ xy + x + y = 11\end{cases}\) или \(\begin{cases}x + y = -9 \\ xy + x + y = 11\end{cases}\)
\(1.\enspace \begin{cases}x = 8 - y \\ 8y - y^{2} + 8 - y + y = 11\end{cases},\)
\(y^{2} - 8y + 3 = 0,\)
\(y_{1{,}2} = \dfrac{8 \pm \sqrt{64 - 12}}{2} = \dfrac{8 \pm 2\sqrt{13}}{2} = 4 \pm \sqrt{13};\)
\(\begin{cases}y = 4 + \sqrt{13} \\ x = 4 - \sqrt{13}\end{cases}\) или \(\begin{cases}y = 4 - \sqrt{13} \\ x = 4 + \sqrt{13}\end{cases}\);
\(2.\enspace \begin{cases}x = -9 - y \\ -9y - y^{2} - 9 - y + y = 11\end{cases}\)
\(\begin{cases}x = -9 - y \\ y^{2} + 9y + 20 = 0\end{cases},\)
\(y_{1{,}2} = \dfrac{-9 \pm \sqrt{81 - 80}}{2},\)
\(y_{1} = -4, y_{2} = -5\)
\(\begin{cases}y = -4 \\ x = -5\end{cases}\) или \(\begin{cases}y = -5 \\ x = -4\end{cases}\).

Ответ: \(\left(2\sqrt{3} - 1; 2\sqrt{3} - 1\right)\), \(\left(-2\sqrt{3} - 1; -2\sqrt{3} - 1\right)\),

\(\left(4 + \sqrt{13}; 4 - \sqrt{13}\right), \left(4 - \sqrt{13}; 4 + \sqrt{13}\right), (-5; -4), (-4; -5);\)

2) \(\begin{cases}\frac{x^{2}}{y} + \frac{y^{2}}{x} = 12 \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{3}\end{cases}\)
\(\begin{cases}\frac{x^{3} + y^{3}}{xy} = 12 \\ \frac{x + y}{xy} = \frac{1}{3}\end{cases},\)
\(\begin{cases}(x + y)\left(x^{2} - xy + y^{2}\right) = 12xy \\ 3(x + y) = xy\end{cases},\)
\(x + y \neq 0\) (иначе по второму уравнению \(xy = 0\), а \(x \neq 0\), \(y \neq 0\)), разделим первое уравнение на \(x + y\):
\(\begin{cases}x^{2} + 2xy + y^{2} = \frac{12xy}{x + y} + 3xy \\ 3(x + y) = xy\end{cases},\)
\(x + y = u, xy = v\)
\(\begin{cases}u^{2} = \frac{12v}{u} + 3v \\ 3u = v\end{cases},\)
\(\begin{cases}v = 3u \\ u^{2} = 36 + 9u\end{cases},\)
\(u^{2} - 9u - 36 = 0,\)
\(u_{1{,}2} = \dfrac{9 \pm \sqrt{81 + 144}}{2},\)
\(u_{1} = 12, u_{2} = -3\)
\(\begin{cases}u = 12 \\ v = 36\end{cases}\) или \(\begin{cases}u = -3 \\ v = -9\end{cases}\)
\(1.\enspace \begin{cases}x + y = 12 \\ xy = 36\end{cases}\)
\(\begin{cases}x = 12 - y \\ 12y - y^{2} = 36\end{cases},\)
\(y^{2} - 12y + 36 = 0,\)
\(y = \dfrac{12 \pm \sqrt{144 - 144}}{2} = 6,\)
\(\begin{cases}y = 6 \\ x = 6\end{cases};\)
\(2.\enspace \begin{cases}x + y = -3 \\ xy = -9\end{cases},\)
\(\begin{cases}x = -3 - y \\ -3y - y^{2} = -9\end{cases},\)
\(y^{2} + 3y - 9 = 0,\)
\(y_{1{,}2} = \dfrac{-3 \pm \sqrt{9 + 36}}{2} = \dfrac{-3 \pm 3\sqrt{5}}{2},\)
\(\begin{cases}y = \frac{-3 + 3\sqrt{5}}{2} \\ x = \frac{-3 - 3\sqrt{5}}{2}\end{cases}\) или \(\begin{cases}y = \frac{-3 - 3\sqrt{5}}{2} \\ x = \frac{-3 + 3\sqrt{5}}{2}\end{cases}\).

Ответ: \((6; 6)\), \(\left(-\frac{3 + 3\sqrt{5}}{2}; \frac{3\sqrt{5} - 3}{2}\right)\), \(\left(\frac{3\sqrt{5} - 3}{2}; -\frac{3 + 3\sqrt{5}}{2}\right)\);

3) \(\begin{cases}2x^{2}y^{2} - 3y^{2} + 5xy - 6 = 0 \\ 3x^{2}y^{2} - 4y^{2} + 3xy - 2 = 0\end{cases}\)

Умножим первое уравнение на \(-4\), а второе на \(3\) и прибавим его к первому:

\(-8x^{2}y^{2} + 12y^{2} - 20xy + 24 + 9x^{2}y^{2} - 12y^{2} + 9xy - 6 = 0,\)
\(x^{2}y^{2} - 11xy + 18 = 0,\)
\(xy = z,\)
\(z^{2} - 11z + 18 = 0,\)
\(z_{1{,}2} = \dfrac{11 \pm \sqrt{121 - 72}}{2},\)
\(z_{1} = 9, z_{2} = 2;\)
\(\begin{cases}xy = 9 \\ 3x^{2}y^{2} - 4y^{2} + 3xy - 2 = 0\end{cases}\) или \(\begin{cases}xy = 2 \\ 3x^{2}y^{2} - 4y^{2} + 3xy - 2 = 0\end{cases}\)
\(1.\enspace \begin{cases}xy = 9 \\ 3 \cdot 81 - 4y^{2} + 27 - 2 = 0\end{cases},\)
\(\begin{cases}xy = 9 \\ -4y^{2} = -268\end{cases},\)
\(\begin{cases}y = \pm \sqrt{67} \\ x = \pm \frac{9}{\sqrt{67}}\end{cases};\)
\(2.\enspace \begin{cases}xy = 2 \\ 3 \cdot 4 - 4y^{2} + 6 - 2 = 0\end{cases},\)
\(\begin{cases}xy = 2 \\ -4y^{2} = -16\end{cases},\)
\(\begin{cases}y = \pm 2 \\ x = \pm 1\end{cases}.\)

Ответ: \((1; 2)\), \((-1; -2)\), \(\left(\frac{9}{\sqrt{67}}; \sqrt{67}\right)\), \(\left(-\frac{9}{\sqrt{67}}; -\sqrt{67}\right)\);

4) \(\begin{cases}\frac{xy}{x + 2y} + \frac{x + 2y}{xy} = 2 \\ \frac{xy}{x - 2y} + \frac{x - 2y}{xy} = 4\end{cases}\)
\(\dfrac{xy}{x + 2y} = u, \dfrac{xy}{x - 2y} = v\)
\(\begin{cases}u + \frac{1}{u} = 2 \\ v + \frac{1}{v} = 4\end{cases},\)
\(\begin{cases}u^{2} - 2u + 1 = 0 \\ v^{2} - 4v + 1 = 0\end{cases},\)
\(\begin{cases}(u - 1)^{2} = 0 \\ v^{2} - 4v + 1 = 0\end{cases},\)
\(v_{1{,}2} = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3},\)
\(\begin{cases}u = 1 \\ v = 2 \pm \sqrt{3}\end{cases};\)
\(\begin{cases}\frac{xy}{x + 2y} = 1 \\ \frac{xy}{x - 2y} = 2 \pm \sqrt{3}\end{cases},\)
\(\begin{cases}x = \frac{2y}{y - 1} \\ \frac{2y^{2}}{(y - 1)\left(\frac{2y}{y - 1} - 2y\right)} = 2 \pm \sqrt{3}\end{cases},\)
\(\begin{cases}x = \frac{2y}{y - 1} \\ \frac{2y^{2}}{2y - 2y^{2} + 2y} = 2 \pm \sqrt{3}\end{cases},\)
\(\begin{cases}x = \frac{2y}{y - 1} \\ \frac{y}{2 - y} = 2 \pm \sqrt{3}\end{cases},\)
\(1.\enspace y = \left(2 + \sqrt{3}\right)(2 - y)\)
\(y = 4 - \sqrt{3}y + 2\sqrt{3} - 2y,\)
\(y\left(1 + \sqrt{3} + 2\right) = 4 + 2\sqrt{3},\)
\(y = \dfrac{4 + 2\sqrt{3}}{3 + \sqrt{3}} = \dfrac{\left(4 + 2\sqrt{3}\right)\left(3 - \sqrt{3}\right)}{9 - 3} =\)
\(= \dfrac{12 - 4\sqrt{3} + 6\sqrt{3} - 6}{6} = 1 + \dfrac{1}{\sqrt{3}};\)
\(2.\enspace y = \left(2 - \sqrt{3}\right)(2 - y)\)
\(y = 4 + \sqrt{3}y - 2\sqrt{3} - 2y,\)
\(y\left(1 - \sqrt{3} + 2\right) = 4 - 2\sqrt{3},\)
\(y = \dfrac{4 - 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{\left(4 - 2\sqrt{3}\right)\left(3 + \sqrt{3}\right)}{9 - 3} =\)
\(= \dfrac{12 - 6\sqrt{3} + 4\sqrt{3} - 6}{6} = 1 - \dfrac{1}{\sqrt{3}};\)
\(\begin{cases}y = 1 + \frac{1}{\sqrt{3}} \\ x = \frac{2 + \frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 2\sqrt{3} + 2\end{cases}\) или \(\begin{cases}y = 1 - \frac{1}{\sqrt{3}} \\ x = \frac{2 - \frac{2}{\sqrt{3}}}{-\frac{1}{\sqrt{3}}} = 2 - 2\sqrt{3}\end{cases}\);

5) \(\begin{cases}x^{3}y + xy^{3} = \frac{10}{9}(x + y)^{2} \\ x^{4}y + xy^{4} = \frac{2}{3}(x + y)^{3}\end{cases}\)
\(\begin{cases}xy\left(x^{2} + y^{2}\right) = \frac{10}{9}(x + y)^{2} \\ xy\left(x^{3} + y^{3}\right) = \frac{2}{3}(x + y)^{3}\end{cases},\)
\(\begin{cases}xy\left(x^{2} + 2xy + y^{2} - 2xy\right) = \frac{10}{9}(x + y)^{2} \\ xy\left(x^{3} + y^{3}\right) = \frac{2}{3}(x + y)^{3}\end{cases},\)
\(\begin{cases}xy\left((x + y)^{2} - 2xy\right) = \frac{10}{9}(x + y)^{2} \\ xy(x + y)\left(x^{2} - xy + y^{2}\right) = \frac{2}{3}(x + y)^{3}\end{cases},\)

Если \(x + y = 0\), то второе уравнение верно, а из первого \(-2x^{2}y^{2} = 0\), т.е. \(x = y = 0\); при \(x + y \neq 0\) разделим второе уравнение на \(x + y\):

\(\begin{cases}xy(x + y)^{2} - 2x^{2}y^{2} - \frac{10}{9}(x + y)^{2} = 0 \\ xy\left(x^{2} + 2xy + y^{2} - 3xy\right) - \frac{2}{3}(x + y)^{2} = 0\end{cases},\)
\(\begin{cases}(x + y)^{2}\left(xy - \frac{10}{9}\right) - 2x^{2}y^{2} = 0 \\ xy(x + y)^{2} - 3x^{2}y^{2} - \frac{2}{3}(x + y)^{2} = 0\end{cases},\)
\(\begin{cases}(x + y)^{2}\left(xy - \frac{10}{9}\right) - 2x^{2}y^{2} = 0 \\ (x + y)^{2}\left(xy - \frac{2}{3}\right) - 3x^{2}y^{2} = 0\end{cases},\)
\(x + y = u, xy = v\)
\(\begin{cases}u^{2}\left(v - \frac{10}{9}\right) - 2v^{2} = 0 \\ u^{2}\left(v - \frac{2}{3}\right) - 3v^{2} = 0\end{cases},\)
\(\begin{cases}u^{2} = \frac{2v^{2}}{v - \frac{10}{9}} = \frac{18v^{2}}{9v - 10} \\ \frac{18v^{2}}{9v - 10}\left(v - \frac{2}{3}\right) - 3v^{2} = 0\end{cases},\)
\(3v^{2} \cdot \dfrac{18v - 12 - 27v + 30}{27v - 30} = 0,\)
\(v_{1} = 0;\)
\(-9v + 18 = 0,\)
\(v_{2} = 2;\)
\(\begin{cases}v = 0 \\ u = 0\end{cases}\) или \(\begin{cases}v = 2 \\ u = \pm 3\end{cases}\)
\(1.\enspace \begin{cases}x + y = 0 \\ xy = 0\end{cases},\)
\(\begin{cases}x = 0 \\ y = 0\end{cases};\)
\(2.\enspace \begin{cases}x + y = 3 \\ xy = 2\end{cases},\)
\(\begin{cases}x = 3 - y \\ 3y - y^{2} - 2 = 0\end{cases},\)
\(y^{2} - 3y + 2 = 0,\)
\(y_{1{,}2} = \dfrac{3 \pm \sqrt{9 - 8}}{2},\)
\(y_{1} = 1, y_{2} = 2\)
\(\begin{cases}y = 1 \\ x = 2\end{cases}\) или \(\begin{cases}y = 2 \\ x = 1\end{cases}\);
\(3.\enspace \begin{cases}x + y = -3 \\ xy = 2\end{cases},\)
\(\begin{cases}x = -3 - y \\ -3y - y^{2} - 2 = 0\end{cases},\)
\(y^{2} + 3y + 2 = 0,\)
\(y_{1{,}2} = \dfrac{-3 \pm \sqrt{9 - 8}}{2},\)
\(y_{1} = -1, y_{2} = -2;\)
\(\begin{cases}y = -1 \\ x = -2\end{cases}\) или \(\begin{cases}y = -2 \\ x = -1\end{cases}\).

Ответ: \((0; 0)\), \((-1; -2)\), \((-2; -1)\), \((1; 2)\), \((2; 1)\);

6) \(\begin{cases}\frac{x\left(y^{2} + 1\right)}{x^{2} + y^{2}} = \frac{3}{5} \\ \frac{y\left(x^{2} - 1\right)}{x^{2} + y^{2}} = \frac{4}{5}\end{cases}\)
\(\left(\dfrac{x\left(y^{2} + 1\right)}{x^{2} + y^{2}}\right)^{2} + \left(\dfrac{y\left(x^{2} - 1\right)}{x^{2} + y^{2}}\right)^{2} = \dfrac{9}{25} + \dfrac{16}{25},\)
\(\dfrac{x^{2}\left(y^{4} + 2y^{2} + 1\right) + y^{2}\left(x^{4} - 2x^{2} + 1\right)}{\left(x^{2} + y^{2}\right)^{2}} = 1,\)
\(\dfrac{x^{2}y^{4} + 2x^{2}y^{2} + x^{2} + y^{2}x^{4} - 2x^{2}y^{2} + y^{2}}{\left(x^{2} + y^{2}\right)^{2}} = 1,\)
\(\dfrac{x^{2}\left(x^{2}y^{2} + 1\right) + y^{2}\left(x^{2}y^{2} + 1\right)}{\left(x^{2} + y^{2}\right)^{2}} = 1,\)
\(\dfrac{\left(x^{2} + y^{2}\right)\left(x^{2}y^{2} + 1\right)}{\left(x^{2} + y^{2}\right)^{2}} = 1,\)
\(x^{2}y^{2} + 1 = x^{2} + y^{2},\)
\(x^{2} + y^{2} - x^{2}y^{2} - 1 = 0,\)
\(-x^{2}\left(y^{2} - 1\right) + y^{2} - 1 = 0,\)
\(\left(1 - x^{2}\right)\left(y^{2} - 1\right) = 0,\)
\(x^{2} \neq 1\), т.к. при этом не выполнится второе уравнение системы.
\(y^{2} = 1,\)
\(y = \pm 1;\)
\(\dfrac{x\left(y^{2} + 1\right)}{x^{2} + y^{2}} = \dfrac{3}{5},\)
\(\dfrac{2x}{x^{2} + 1} = \dfrac{3}{5},\)
\(10x = 3x^{2} + 3,\)
\(3x^{2} + 3 - 10x = 0,\)
\(x_{1{,}2} = \dfrac{10 \pm \sqrt{100 - 36}}{6},\)
\(x_{1} = 3, x_{2} = \dfrac{1}{3}.\)

Из второго уравнения при \(x = 3\): \(\frac{8y}{10} = \frac{4}{5}\), \(y = 1\); при \(x = \frac{1}{3}\): \(-\frac{8y}{10} = \frac{4}{5}\), \(y = -1\).

Ответ: \((3; 1)\), \(\left(\frac{1}{3}; -1\right)\).