Страница 228 номер 784, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(\frac{x^{2}}{3} + \frac{48}{x^{2}} = 10\left(\frac{x}{3} - \frac{4}{x}\right)\)
\(z = \dfrac{x}{3} - \dfrac{4}{x},\)
\(z^{2} = \dfrac{x^{2}}{9} - \dfrac{8}{3} + \dfrac{16}{x^{2}},\)
\(3z^{2} = \dfrac{x^{2}}{3} - 8 + \dfrac{48}{x^{2}},\)
\(3z^{2} + 8 = 10z,\)
\(3z^{2} - 10z + 8 = 0,\)
\(z_{1{,}2} = \dfrac{10 \pm \sqrt{100 - 96}}{6},\)
\(z_{1} = 2, z_{2} = \dfrac{4}{3};\)
\(\frac{x}{3} - \frac{4}{x} = 2\) или \(\frac{x}{3} - \frac{4}{x} = \frac{4}{3}\)
\(1.\enspace x^{2} - 12 - 6x = 0,\)
\(x_{1{,}2} = \dfrac{6 \pm \sqrt{36 + 48}}{2} = 3 \pm \sqrt{21};\)
\(2.\enspace x^{2} - 12 - 4x = 0,\)
\(x_{3{,}4} = \dfrac{4 \pm \sqrt{16 + 48}}{2},\)
\(x_{3} = 6, x_{4} = -2.\)

Ответ: \(x_{1{,}2} = 3 \pm \sqrt{21}\), \(x_{3} = 6\), \(x_{4} = -2\);

2) \(6x^{4} - 35x^{3} + 62x^{2} - 35x + 6 = 0\)
\(x = 0\) не является корнем (\(6 \neq 0\)), разделим обе части на \(x^{2}\):
\(6x^{2} - 35x + 62 - \dfrac{35}{x} + \dfrac{6}{x^{2}} = 0,\)
\(6\left(x^{2} + \dfrac{1}{x^{2}}\right) - 35\left(x + \dfrac{1}{x}\right) + 62 = 0,\)
\(z = x + \dfrac{1}{x},\)
\(z^{2} = x^{2} + 2 + \dfrac{1}{x^{2}},\)
\(x^{2} + \dfrac{1}{x^{2}} = z^{2} - 2,\)
\(6z^{2} - 12 - 35z + 62 = 0,\)
\(6z^{2} - 35z + 50 = 0,\)
\(z_{1{,}2} = \dfrac{35 \pm \sqrt{1225 - 1200}}{12},\)
\(z_{1} = \dfrac{10}{3}, z_{2} = \dfrac{5}{2};\)
\(x + \frac{1}{x} = \frac{10}{3}\) или \(x + \frac{1}{x} = \frac{5}{2}\)
\(1.\enspace 3x^{2} - 10x + 3 = 0,\)
\(x_{1{,}2} = \dfrac{10 \pm \sqrt{100 - 36}}{6},\)
\(x_{1} = 3, x_{2} = \dfrac{1}{3};\)
\(2.\enspace 2x^{2} - 5x + 2 = 0,\)
\(x_{3{,}4} = \dfrac{5 \pm \sqrt{25 - 16}}{4},\)
\(x_{3} = 2, x_{4} = \dfrac{1}{2};\)

3) \(\sqrt{x - 2 + \sqrt{2x - 5}} + \sqrt{x + 2 + 3\sqrt{2x - 5}} = 7\sqrt{2}\)
\(\sqrt{2x - 5} = 2z, z \geqslant 0,\)
\(2x - 5 = 4z^{2},\)
\(x = 2z^{2} + \dfrac{5}{2},\)
\(\sqrt{2z^{2} + \dfrac{5}{2} - 2 + 2z} + \sqrt{2z^{2} + \dfrac{5}{2} + 2 + 6z} = 7\sqrt{2},\)
\(\sqrt{\dfrac{4z^{2} + 5 - 4 + 4z}{2}} + \sqrt{\dfrac{4z^{2} + 5 + 4 + 12z}{2}} = 7\sqrt{2},\)
\(\sqrt{(2z + 1)^{2}} + \sqrt{(2z + 3)^{2}} = 14,\)
\(2z + 1 + 2z + 3 = 14,\)
\(4z = 10,\)
\(z = \dfrac{5}{2};\)
\(\sqrt{2x - 5} = 5,\)
\(2x - 5 = 25,\)
\(x = 15;\)

4) \(2\sqrt{x^{2} - 2x + 4} - \sqrt{x^{2} - 2x + 9} = 1\)
\(x^{2} - 2x + 4 = z,\)
\(x^{2} - 2x + 9 = z + 5,\)
\(2\sqrt{z} - \sqrt{z + 5} = 1,\)
\(4z - 4\sqrt{z^{2} + 5z} + z + 5 = 1,\)
\(5z + 4 = 4\sqrt{z^{2} + 5z},\)
\(25z^{2} + 40z + 16 = 16z^{2} + 80z,\)
\(9z^{2} - 40z + 16 = 0,\)
\(z_{1{,}2} = \dfrac{40 \pm \sqrt{1600 - 576}}{18},\)
\(z_{1} = \dfrac{4}{9}, z_{2} = 4;\)

Проверка:

\(2\sqrt{\dfrac{4}{9}} - \sqrt{\dfrac{4}{9} + 5} = 1,\)
\(2 \cdot \dfrac{2}{3} - \dfrac{7}{3} = 1,\)
\(-1 = 1\) – не верно. \(\frac{4}{9}\) – не корень;
\(2\sqrt{4} - \sqrt{4 + 5} = 1,\)
\(4 - 3 = 1\) – верно;
\(x^{2} - 2x + 4 = 4,\)
\(x(x - 2) = 0,\)
\(x_{1} = 0, x_{2} = 2.\)