第8章 8.20

(1)

$\frac{x}{x+1} = \frac{5}{x-1} \qquad x(x-1) = 5(x+1) \qquad x^2 - x = 5x + 5 \qquad x^2 - 6x - 5 = 0 \qquad x = 3 \pm \sqrt{14}$

(2)

$\frac{x^2}{x+2} + \frac{2x+4}{x^2} = 3 \qquad \frac{x^4 + (2x+4)(x+2)}{x^2(x+2)} = 3 \qquad x^4 + 2x^2 + 8x + 8 = 3x^3 + 6x^2$

$x^4 - 3x^3 - 4x^2 + 8x + 8 = 0 \qquad (x+1)(x-2)(x^2 - 2x - 4) = 0$

$x = -1, 2, 1 \pm \sqrt{5}$

(3)

$\frac{x-1}{x+4} + \frac{x-3}{x+2} + \frac{x-5}{x} = 3 \qquad \frac{(x-1)x(x+2) + (x-3)x(x+4) + (x-5)(x+4)(x+2)}{x(x+2)(x+4)} = 3$

$3x^3 + 3x^2 - 36x - 40 = 3x^3 + 18x^2 + 24x$

$15x^2 + 60x + 40 = 0 \qquad 3x^2 + 12x + 8 = 0 \qquad x = \frac{-6 \pm 2\sqrt{3}}{3}$

(4)

$\frac{2x}{3x^2+1} + \frac{3x^2+1}{2x} = \frac{5}{2} \qquad \frac{4x^2 + (3x^2+1)^2}{2x(3x^2+1)} = \frac{5}{2}$

$2\{4x^2 + 9x^4 + 6x^2 + 1\} = 5(6x^3 + 2x)$

$18x^4 - 30x^3 + 20x^2 - 10x + 2 = 0 \qquad 9x^4 - 15x^3 + 10x^2 - 5x + 1 = 0$

$(x-1)(3x-1)(3x^2 - x + 1) = 0 \qquad x = 1, \frac{1}{3}, \frac{1 \pm \sqrt{11}i}{6}$

(5)

$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$

$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} = -\frac{1}{x^2-13x-8} \qquad \frac{x^2+2x-8+x^2+11x-8}{(x^2+11x-8)(x^2+2x-8)} = -\frac{1}{x^2-13x-8}$

$(2x^2+13x-16)(x^2-13x-8) = -(x^2+11x-8)(x^2+2x-8)$

$2x^4 - 13x^3 - 201x^2 + 104x + 128 = -x^4 - 13x^3 - 6x^2 + 104x - 64$

$3(x^4 - 65x^2 + 64) = 3(x^2-1)(x^2-64) = 0 \qquad x = \pm 1, \pm 8$