696. Найдите корни уравнения: a) x√3 + √2 x√3-√2 10x + = x√3-√2 x√3 + √2 3x²-2 ; б) 1-y√5 1 + y√5 9y + = 1+ y√5 1-y√5 1-5 y² .

Ответ:

696.

а)

\[\frac{x\sqrt{3} + \sqrt{2}}{x\sqrt{3} - \sqrt{2}} + \frac{x\sqrt{3} - \sqrt{2}}{x\sqrt{3} + \sqrt{2}} = \frac{10x}{3x^2 - 2}\]

\[\frac{(x\sqrt{3} + \sqrt{2})^2 + (x\sqrt{3} - \sqrt{2})^2}{(x\sqrt{3} - \sqrt{2})(x\sqrt{3} + \sqrt{2})} = \frac{10x}{3x^2 - 2}\]

\[\frac{3x^2 + 2x\sqrt{6} + 2 + 3x^2 - 2x\sqrt{6} + 2}{3x^2 - 2} = \frac{10x}{3x^2 - 2}\]

\[\frac{6x^2 + 4}{3x^2 - 2} = \frac{10x}{3x^2 - 2}\]

\[6x^2 + 4 = 10x\]

\[6x^2 - 10x + 4 = 0\]

\[3x^2 - 5x + 2 = 0\]

\[D = (-5)^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1\]

\[x = \frac{5 \pm \sqrt{1}}{6} = \frac{5 \pm 1}{6}\]

\[x_1 = \frac{5 + 1}{6} = 1, x_2 = \frac{5 - 1}{6} = \frac{2}{3}\]

Ответ: \[x_1 = 1, x_2 = \frac{2}{3}\]

б)

\[\frac{1 - y\sqrt{5}}{1 + y\sqrt{5}} + \frac{1 + y\sqrt{5}}{1 - y\sqrt{5}} = \frac{9y}{1 - 5y^2}\]

\[\frac{(1 - y\sqrt{5})^2 + (1 + y\sqrt{5})^2}{(1 + y\sqrt{5})(1 - y\sqrt{5})} = \frac{9y}{1 - 5y^2}\]

\[\frac{1 - 2y\sqrt{5} + 5y^2 + 1 + 2y\sqrt{5} + 5y^2}{1 - 5y^2} = \frac{9y}{1 - 5y^2}\]

\[\frac{2 + 10y^2}{1 - 5y^2} = \frac{9y}{1 - 5y^2}\]

\[2 + 10y^2 = 9y\]

\[10y^2 - 9y + 2 = 0\]

\[D = (-9)^2 - 4 \cdot 10 \cdot 2 = 81 - 80 = 1\]

\[y = \frac{9 \pm \sqrt{1}}{20} = \frac{9 \pm 1}{20}\]

\[y_1 = \frac{9 + 1}{20} = \frac{1}{2}, y_2 = \frac{9 - 1}{20} = \frac{2}{5}\]

Ответ: \[y_1 = \frac{1}{2}, y_2 = \frac{2}{5}\]