695. Решите уравнение: a) 2x + 1 3(2x - 1) 8 1- 4x2 = 0; = 2x - 1 + 7(2x + 1) б) y y2 - 9 y2 + 3y 6y + 2y2 = 0;

Ответ:

695.

а)

\[\frac{2x + 1}{2x - 1} + \frac{3(2x - 1)}{7(2x + 1)} + \frac{8}{1 - 4x^2} = 0\]

\[\frac{2x + 1}{2x - 1} + \frac{3(2x - 1)}{7(2x + 1)} - \frac{8}{4x^2 - 1} = 0\]

\[\frac{2x + 1}{2x - 1} + \frac{3(2x - 1)}{7(2x + 1)} - \frac{8}{(2x - 1)(2x + 1)} = 0\]

Общий знаменатель: \[(2x - 1)(2x + 1) \cdot 7\]

\[\frac{7(2x + 1)^2 + 3(2x - 1)^2 - 8 \cdot 7}{7(2x - 1)(2x + 1)} = 0\]

\[7(4x^2 + 4x + 1) + 3(4x^2 - 4x + 1) - 56 = 0\]

\[28x^2 + 28x + 7 + 12x^2 - 12x + 3 - 56 = 0\]

\[40x^2 + 16x - 46 = 0\]

\[20x^2 + 8x - 23 = 0\]

\[D = 8^2 - 4 \cdot 20 \cdot (-23) = 64 + 1840 = 1904\]

\[x = \frac{-8 \pm \sqrt{1904}}{40} = \frac{-8 \pm 4\sqrt{119}}{40} = \frac{-2 \pm \sqrt{119}}{10}\]

Ответ: \[x = \frac{-2 \pm \sqrt{119}}{10}\]

б)

\[\frac{y}{y^2 - 9} + \frac{1}{y^2 + 3y} + \frac{6y}{6y + 2y^2} = 0\]

\[\frac{y}{(y - 3)(y + 3)} + \frac{1}{y(y + 3)} + \frac{6y}{2y(3 + y)} = 0\]

\[\frac{y}{(y - 3)(y + 3)} + \frac{1}{y(y + 3)} + \frac{3}{3 + y} = 0\]

Общий знаменатель: \[(y - 3)(y + 3)y\]

\[\frac{y^2 + (y - 3) + 3y(y - 3)}{y(y - 3)(y + 3)} = 0\]

\[y^2 + y - 3 + 3y^2 - 9y = 0\]

\[4y^2 - 8y - 3 = 0\]

\[D = (-8)^2 - 4 \cdot 4 \cdot (-3) = 64 + 48 = 112\]

\[y = \frac{8 \pm \sqrt{112}}{8} = \frac{8 \pm 4\sqrt{7}}{8} = \frac{2 \pm \sqrt{7}}{2}\]

Ответ: \[y = \frac{2 \pm \sqrt{7}}{2}\]