Страница 62 номер 156, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sqrt{2x - 34} = 1 + \sqrt{x}\), \(\left(\sqrt{2x - 34}\right)^{2} = \left(1 + \sqrt{x}\right)^{2}\),
\(2x - 34 = 1 + 2\sqrt{x} + x, x - 35 = 2\sqrt{x}, (x - 35)^{2} = \left(2\sqrt{x}\right)^{2},\)
\(x^{2} - 70x + 1225 - 4x = 0, x^{2} - 74x + 1225 = 0,\)
\(x_{1{,}2} = \dfrac{74 \pm \sqrt{5476 - 4900}}{2},\)
\(x_{1} = 49, x_{2} = 25.\)

Проверка: \(\sqrt{98 - 34} = 1 + \sqrt{49}\) – верно; \(\sqrt{50 - 34} = 1 + \sqrt{25}\) – неверно.

Ответ: \(x = 49\).

2) \(\sqrt{5x} + \sqrt{14 - x} = 8\), \(\left(\sqrt{5x} + \sqrt{14 - x}\right)^{2} = 8^{2}\),
\(5x + 2\sqrt{70x - 5x^{2}} + 14 - x - 64 = 0,\)
\((25 - 2x)^{2} = \left(\sqrt{70x - 5x^{2}}\right)^{2},\)
\(625 - 100x + 4x^{2} - 70x + 5x^{2} = 0, 9x^{2} - 170x + 625 = 0,\)
\(x_{1{,}2} = \dfrac{170 \pm \sqrt{28900 - 22500}}{18}, x_{1} = \dfrac{250}{18} = \dfrac{125}{9} = 13\dfrac{8}{9}, x_{2} = 5.\)

Проверка: \(\sqrt{5 \cdot 5} + \sqrt{14 - 5} = 8\) – верно; \(\sqrt{\frac{625}{9}} + \sqrt{\frac{126 - 125}{9}} = 8\), \(\frac{25}{3} + \frac{1}{3} = 8\) – неверно.

Ответ: \(x = 5\).

3) \(\sqrt{15 + x} + \sqrt{3 + x} = 6\), \(\left(\sqrt{15 + x} + \sqrt{3 + x}\right)^{2} = 6^{2}\),
\(15 + x + 2\sqrt{(15 + x)(3 + x)} + 3 + x = 36,\)
\((9 - x)^{2} = \left(\sqrt{45 + 18x + x^{2}}\right)^{2},\)
\(81 - 18x + x^{2} - 45 - 18x - x^{2} = 0,\)
\(36 - 36x = 0, x = 1.\)

Проверка: \(\sqrt{15 + 1} + \sqrt{3 + 1} = 6\) – верно.

4) \(\sqrt{3 - 2x} - \sqrt{1 - x} = 1\), \(\left(\sqrt{3 - 2x} - \sqrt{1 - x}\right)^{2} = 1\),
\(3 - 2x - 2\sqrt{(3 - 2x)(1 - x)} + 1 - x - 1 = 0,\)
\((3x - 3)^{2} = 4\left(\sqrt{3 - 5x + 2x^{2}}\right)^{2},\)
\(9x^{2} - 18x + 9 - 12 + 20x - 8x^{2} = 0,\)
\(x^{2} + 2x - 3 = 0, x_{1{,}2} = \dfrac{-2 \pm \sqrt{4 + 12}}{2}, x_{1} = 1, x_{2} = -3.\)

Проверка: \(\sqrt{3 - 2} - \sqrt{1 - 1} = 1\) – верно; \(\sqrt{3 + 6} - \sqrt{1 + 3} = 1\) – верно.

Ответ: \(x_{1} = 1\), \(x_{2} = -3\).