Страница 70 номер 183, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sqrt{3 - x} = 2\), \(\left( \sqrt{3 - x} \right)^{2} = 2^{2}\), \(3 - x = 4\), \(x = -1\);

2) \(\sqrt{3x + 1} = 8\), \(\left( \sqrt{3x + 1} \right)^{2} = 8^{2}\), \(3x + 1 = 64\), \(3x = 63\), \(x = 21\);

3) \(\sqrt{3 - 4x} = 2x\), \(\left( \sqrt{3 - 4x} \right)^{2} = (2x)^{2}\), \(3 - 4x - 4x^{2} = 0\),
\(4x^{2} + 4x - 3 = 0,\ x_{1{,}2} = \dfrac{-4 \pm \sqrt{16 + 48}}{8},\ x_{1} = \dfrac{1}{2},\ x_{2} = -1{,}5.\)

Проверка: \(\sqrt{3 - 4 \cdot \frac{1}{2}} = 2 \cdot \frac{1}{2}\) – верно; \(\sqrt{3 + 4 \cdot 1{,}5} = -2 \cdot 1{,}5\) – неверно.

Ответ: \(x = \frac{1}{2}\).

4) \(\sqrt{5x - 1 + 3x^{2}} = 3x\), \(\left( \sqrt{5x - 1 + 3x^{2}} \right)^{2} = (3x)^{2}\), \(5x - 1 - 6x^{2} = 0\),
\(6x^{2} - 5x + 1 = 0,\ x_{1{,}2} = \dfrac{5 \pm \sqrt{25 - 24}}{12},\ x_{1} = \dfrac{1}{2},\ x_{2} = \dfrac{1}{3}.\)

Проверка: \(\sqrt{5 \cdot \frac{1}{2} - 1 + 3 \cdot \frac{1}{4}} = 3 \cdot \frac{1}{2}\) – верно;

\(\sqrt{5 \cdot \frac{1}{3} - 1 + 3 \cdot \frac{1}{9}} = 3 \cdot \frac{1}{3}\) – верно.

5) \(\sqrt[3]{x^{2} - 17} = 2\), \(x^{2} - 17 = 8\), \(x^{2} = 25\), \(x_{1{,}2} = \pm 5\);

6) \(\sqrt[4]{x^{2} + 17} = 3\), \(x^{2} + 17 = 81\), \(x^{2} = 64\), \(x_{1{,}2} = \pm 8\).