ГДЗ по Алгебре 9 Класс Номер 13.11 Углубленный Уровень Мерзляк, Поляков — Подробные Ответы

Решите систему уравнений:
1) \(x^2 + 3xy = 7, y^2 + xy = 6\)
2) \(x^2 + xy — y^2 = 20, x^2 + 3xy — 3y^2 = 28\)
3) \(x^2 + 4xy — 3y^2 = 2, x^2 — xy + 5y^2 = 5\)
4) \(2x^2 — 3xy + 2y^2 = 14, x^2 + xy — y^2 = 5\)

1)
\((x^2 + 3xy = 7) \cdot 6\)
\((y^2 + xy = 6) \cdot 7\)
\(\Rightarrow \begin{cases} 6x^2 + 18xy = 42 \\ 7y^2 + 7xy = 42 \end{cases}\)
\(6x^2 + 11xy — 7y^2 = 0\);
\(D = (11y)^2 + 4 \cdot 6 \cdot 7y^2 = 121y^2 + 168y^2 = 289y^2\), тогда:
\(x_1 = \frac{-11y — 17y}{2 \cdot 6} = \frac{-28y}{12} = \frac{-7y}{3}\) и \(x_2 = \frac{-11y + 17y}{2 \cdot 6} = \frac{6y}{12} = \frac{y}{2}\).

Первое значение:
\(y^2 + y \cdot \left(-\frac{7y}{3}\right) = 6 \cdot 3\);
\(3y^2 — 7y^2 = 18\);
\(4y^2 = -18\);
\(y \in \emptyset\).

Второе значение:
\(y^2 + y \cdot \frac{y}{2} = 6 \cdot 2\);
\(2y^2 + y^2 = 12\);
\(3y^2 = 12\);
\(y^2 = 4\);
\(y = \pm 2\);
\(x = \pm 1\);

Ответ: \((-1; -2); (1; 2)\).

2)
\((x^2 + xy — y^2 = 20) \cdot 7\)
\((x^2 + 3xy — 3y^2 = 28) \cdot 5\)
\(\Rightarrow \begin{cases} 7x^2 + 7xy — 7y^2 = 140 \\ 5x^2 + 15xy — 15y^2 = 140 \end{cases}\)
\(2x^2 — 8xy + 8y^2 = 0 \cdot 2\);
\(x^2 — 4xy + 4y^2 = 0\);
\((x — 2y)^2 = 0\);
\(x = 2y\).

Из первого уравнения:
\((2y)^2 + y \cdot 2y — y^2 = 20\);
\(4y^2 + 2y^2 — y^2 = 20\);
\(5y^2 = 20\);
\(y^2 = 4\);
\(y = \pm 2\);
\(x = 2 \cdot (\pm 2) = \pm 4\).

Ответ: \((-4; -2); (4; 2)\).

3)
\((x^2 + 4xy — 3y^2 = 2) \cdot 5\)
\((x^2 — xy + 5y^2 = 5) \cdot 2\)
\(\Rightarrow \begin{cases} 5x^2 + 20xy — 15y^2 = 10 \\ 2x^2 — 2xy + 10y^2 = 10 \end{cases}\)
\(3x^2 + 22xy — 25y^2 = 0\);
\(D = (22y)^2 + 4 \cdot 3 \cdot 25y^2 = 484y^2 + 300y^2 = 784y^2\), тогда:
\(x_1 = \frac{-22y — 28y}{2 \cdot 3} = \frac{-50y}{6} = \frac{-25y}{3}\);
\(x_2 = \frac{-22y + 28y}{2 \cdot 3} = \frac{6y}{6} = y\).

Первое значение:
\(\left(\frac{-25y}{3}\right)^2 + y \cdot \left(\frac{-25y}{3}\right) + 5y^2 = 5 \cdot 9\);
\(\frac{625y^2}{9} + \frac{-75y^2}{9} + \frac{45y^2}{9} = \frac{45}{9}\);
\(\frac{745y^2}{9} = \frac{45}{9}\);
\(y^2 = \frac{45}{745} = \frac{9}{149}\);
\(y = \pm \frac{\sqrt{149}}{149};\)
\(x = \frac{-25}{3} \cdot \left(\pm \frac{\sqrt{149}}{149}\right) = \pm \frac{25\sqrt{149}}{447}\).

Второе значение:
\(y^2 — y^2 + 5y^2 = 5\);
\(5y^2 = 5;\)
\(y^2 = 1;\)
\(x = \pm 1\).

Ответ: \(\left(-\frac{25\sqrt{149}}{447}; -\frac{\sqrt{149}}{149}\right); \left(\frac{25\sqrt{149}}{447}; \frac{\sqrt{149}}{149}\right); (-1; -1); (1; 1)\).

4)
\((2x^2 — 3xy + 2y^2 = 14) \cdot 5\)
\((x^2 + xy — y^2 = 5) \cdot 14\)
\(\Rightarrow \begin{cases} 10x^2 — 15xy + 10y^2 = 70 \\ 14x^2 + 14xy — 14y^2 = 70 \end{cases}\)
\(-4x^2 — 29xy + 24y^2 = 0\);
\(4x^2 + 29xy — 24y^2 = 0\);
\(D = (29y)^2 + 4 \cdot 4 \cdot 24y^2 = 841y^2 + 384y^2 = 1225y^2\), тогда:
\(x_1 = \frac{-29y — 35y}{2 \cdot 4} = \frac{-64y}{8} = -8y;\)
\(x_2 = \frac{-29y + 35y}{2 \cdot 4} = \frac{6y}{8} = \frac{3y}{4}\).

Первое значение:
\((-8y)^2 + y \cdot (-8y) — y^2 = 5;\)
\(64y^2 — 8y^2 — y^2 = 5;\)
\(55y^2 = 5;\)
\(y^2 = \frac{5}{55} = \frac{1}{11};\)
\(y = \pm \frac{1}{\sqrt{11}};\)
\(x = -8 \cdot \left(\pm \frac{1}{\sqrt{11}}\right) = \pm \frac{-8}{\sqrt{11}}.\)

Второе значение:
\(\left(\frac{3y}{4}\right)^2 + y \cdot \frac{3y}{4} — y^2 = 5 \cdot 16;\)
\(\frac{9y^2}{16} + \frac{12y^2}{16} — \frac{16y^2}{16} = \frac{80}{16};\)
\(\frac{5y^2}{16} = \frac{80}{16};\)
\(y^2 = 16;\)
\(y = \pm 4;\)
\(x = \frac{3}{4} \cdot (\pm 4) = \pm 3.\)

Ответ: \(\left(\frac{-8}{\sqrt{11}}; \frac{-1}{\sqrt{11}}\right); \left(\frac{-8}{\sqrt{11}}; \frac{1}{\sqrt{11}}\right); (-3; -4); (3; 4)\).

1)
\((x^2 + 3xy = 7) \cdot 6\)
\((y^2 + xy = 6) \cdot 7\)
\(\Rightarrow \begin{cases} 6x^2 + 18xy = 42 \\ 7y^2 + 7xy = 42 \end{cases}\)
\(6x^2 + 11xy — 7y^2 = 0\)

\(D = (11y)^2 + 4 \cdot 6 \cdot (-7y^2) = 121y^2 + 168y^2 = 289y^2\), тогда:
\(x_1 = \frac{-11y — 17y}{2 \cdot 6} = \frac{-28y}{12} = \frac{-7y}{3}\)
\(x_2 = \frac{-11y + 17y}{2 \cdot 6} = \frac{6y}{12} = \frac{y}{2}\)

Первое значение:
\(y^2 + y \cdot \left(-\frac{7y}{3}\right) = 6 \cdot 3\)
\(3y^2 — 7y^2 = 18\)
\(4y^2 = -18\)
\(y \in \emptyset\)

Второе значение:
\(y^2 + y \cdot \frac{y}{2} = 6 \cdot 2\)
\(2y^2 + y^2 = 12\)
\(3y^2 = 12\)
\(y^2 = 4\)
\(y = \pm 2\)
\(x = \pm 1\)

Ответ: \((-1; -2); (1; 2)\)

2)
\((x^2 + xy — y^2 = 20) \cdot 7\)
\((x^2 + 3xy — 3y^2 = 28) \cdot 5\)
\(\Rightarrow \begin{cases} 7x^2 + 7xy — 7y^2 = 140 \\ 5x^2 + 15xy — 15y^2 = 140 \end{cases}\)
\(2x^2 — 8xy + 8y^2 = 0 \cdot 2\)
\(x^2 — 4xy + 4y^2 = 0\)
\((x — 2y)^2 = 0\)
\(x = 2y\)

Из первого уравнения:
\((2y)^2 + y \cdot 2y — y^2 = 20\)
\(4y^2 + 2y^2 — y^2 = 20\)
\(5y^2 = 20\)
\(y^2 = 4\)
\(y = \pm 2\)
\(x = 2 \cdot (\pm 2) = \pm 4\)

Ответ: \((-4; -2); (4; 2)\)

3)
\((x^2 + 4xy — 3y^2 = 2) \cdot 5\)
\((x^2 — xy + 5y^2 = 5) \cdot 2\)
\(\Rightarrow \begin{cases} 5x^2 + 20xy — 15y^2 = 10 \\ 2x^2 — 2xy + 10y^2 = 10 \end{cases}\)
\(3x^2 + 22xy — 25y^2 = 0\)

\(D = (22y)^2 + 4 \cdot 3 \cdot (-25y^2) = 484y^2 + 300y^2 = 784y^2\), тогда:
\(x_1 = \frac{-22y — 28y}{2 \cdot 3} = \frac{-50y}{6} = \frac{-25y}{3}\)
\(x_2 = \frac{-22y + 28y}{2 \cdot 3} = \frac{6y}{6} = y\)

Первое значение:
\(\left(\frac{-25y}{3}\right)^2 + y \cdot \left(\frac{-25y}{3}\right) + 5y^2 = 5 \cdot 9\)
\(\frac{625y^2}{9} + \frac{-75y^2}{9} + \frac{45y^2}{9} = \frac{45}{9}\)
\(\frac{745y^2}{9} = \frac{45}{9}\)
\(y^2 = \frac{45}{745} = \frac{9}{149}\)
\(y = \pm \frac{\sqrt{149}}{149}\)
\(x = \frac{-25}{3} \cdot \left(\pm \frac{\sqrt{149}}{149}\right) = \pm \frac{25\sqrt{149}}{447}\)

Второе значение:
\(y^2 — y^2 + 5y^2 = 5\)
\(5y^2 = 5\)
\(y^2 = 1\)
\(x = \pm 1\)

Ответ: \(\left(-\frac{25\sqrt{149}}{447}; -\frac{\sqrt{149}}{149}\right); \left(\frac{25\sqrt{149}}{447}; \frac{\sqrt{149}}{149}\right); (-1; -1); (1; 1)\)

4)
\((2x^2 — 3xy + 2y^2 = 14) \cdot 5\)
\((x^2 + xy — y^2 = 5) \cdot 14\)
\(\Rightarrow \begin{cases} 10x^2 — 15xy + 10y^2 = 70 \\ 14x^2 + 14xy — 14y^2 = 70 \end{cases}\)
\(-4x^2 — 29xy + 24y^2 = 0\)
\(4x^2 + 29xy — 24y^2 = 0\)

\(D = (29y)^2 + 4 \cdot 4 \cdot (-24y^2) = 841y^2 + 384y^2 = 1225y^2\), тогда:
\(x_1 = \frac{-29y — 35y}{2 \cdot 4} = \frac{-64y}{8} = -8y\)
\(x_2 = \frac{-29y + 35y}{2 \cdot 4} = \frac{6y}{8} = \frac{3y}{4}\)

Первое значение:
\((-8y)^2 + y \cdot (-8y) — y^2 = 5\)
\(64y^2 — 8y^2 — y^2 = 5\)
\(55y^2 = 5\)
\(y^2 = \frac{5}{55} = \frac{1}{11}\)
\(y = \pm \frac{1}{\sqrt{11}}\)
\(x = -8 \cdot \left(\pm \frac{1}{\sqrt{11}}\right) = \pm \frac{-8}{\sqrt{11}}\)

Второе значение:
\(\left(\frac{3y}{4}\right)^2 + y \cdot \frac{3y}{4} — y^2 = 5 \cdot 16\)
\(\frac{9y^2}{16} + \frac{12y^2}{16} — \frac{16y^2}{16} = \frac{80}{16}\)
\(\frac{5y^2}{16} = \frac{80}{16}\)
\(y^2 = 16\)
\(y = \pm 4\)
\(x = \frac{3}{4} \cdot (\pm 4) = \pm 3\)

Ответ: \(\left(\frac{-8}{\sqrt{11}}; \frac{-1}{\sqrt{11}}\right); \left(\frac{-8}{\sqrt{11}}; \frac{1}{\sqrt{11}}\right); (-3; -4); (3; 4)\)