1+x/(x-3)=-1/(x^2-3x)

解答

1 + \frac{x}{x - 3} = - \frac{1}{x ^{2} - 3 x}

x = 1, x = \frac{1}{2}

十进制

x = 1, x = 0.5

求解步骤

1 + \frac{x}{x - 3} = - \frac{1}{x ^{2} - 3 x}

乘以最小公倍数

x (x - 3) + x^{2} = - 1

解

x (x - 3) + x^{2} = - 1 : x = 1, x = \frac{1}{2}

x = 1, x = \frac{1}{2}

验证解

找到无定义的点（奇点）:

x = 3, x = 0

将不在定义域的点与解相综合:

x = 1, x = \frac{1}{2}

so l v e f or y, 4 y = \frac{6 + x ^{5}}{x ^{5} y}

\frac{x}{x + 1} - \frac{x - 1}{x} = \frac{1}{20}

so l v e f or h, h = 5 - \frac{k}{7 + k}

\frac{x}{x + 2} - 5 = - \frac{2}{x + 2}

\frac{2 t + 1}{2 t + 3} = \frac{3 t - 1}{3 t + 4}