Страница 63 номер 160, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sqrt[3]{x - 2} = 2\), \(\left(\sqrt[3]{x - 2}\right)^{3} = 2^{3}\), \(x - 2 = 8\), \(x = 10\);

2) \(\sqrt[3]{2x + 7} = \sqrt[3]{3(x - 1)}\),
\(\left(\sqrt[3]{2x + 7}\right)^{3} = \left(\sqrt[3]{3(x - 1)}\right)^{3}\), \(2x + 7 = 3x - 3\)
\(x = 10;\)

3) \(\sqrt[4]{25x^{2} - 144} = x\), \(\left(\sqrt[4]{25x^{2} - 144}\right)^{4} = x^{4}\), \(25x^{2} - 144 - x^{4} = 0\),
\(x^{2} = z, z^{2} - 25z + 144 = 0,\)
\(z_{1{,}2} = \dfrac{25 \pm \sqrt{625 - 576}}{2}, z_{1} = 16, z_{2} = 9.\)
\(x_{1} = 4, x_{2} = -4, x_{3} = 3, x_{4} = -3.\)

Проверка: \(\sqrt[4]{25 \cdot 16 - 144} = 4\) – верно; \(\sqrt[4]{25 \cdot 16 - 144} = -4\) – неверно; \(\sqrt[4]{25 \cdot 9 - 144} = 3\) – верно; \(\sqrt[4]{25 \cdot 9 - 144} = -3\) – неверно.

Ответ: \(x_{1} = 4\), \(x_{2} = 3\).

4) \(x^{2} = \sqrt{19x^{2} - 34}\), \((x^{2})^{2} = \left(\sqrt{19x^{2} - 34}\right)^{2}\), \(x^{4} - 19x^{2} + 34 = 0\),
\(x^{2} = z, z^{2} - 19z + 34 = 0,\)
\(z_{1{,}2} = \dfrac{19 \pm \sqrt{361 - 136}}{2}, z_{1} = 17, z_{2} = 2.\)
\(x_{1{,}2} = \pm\sqrt{17}, x_{3{,}4} = \pm\sqrt{2}.\)

Проверка: \(17 = \sqrt{19 \cdot 17 - 34}\) – верно; \(17 = \sqrt{19 \cdot 17 - 34}\) – верно; \(2 = \sqrt{19 \cdot 2 - 34}\) – верно; \(2 = \sqrt{19 \cdot 2 - 34}\) – верно.

Ответ: \(x_{1{,}2} = \pm\sqrt{17}\), \(x_{3{,}4} = \pm\sqrt{2}\).