(4x^3-1)/(2x^2+x-1)=0

解答

\frac{4 x ^{3} - 1}{2 x ^{2} + x - 1} = 0

x = \frac{1}{3 4}, x = - \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = - \frac{1}{2 3 4} - \frac{3}{2 3 4} i

求解步骤

\frac{4 x ^{3} - 1}{2 x ^{2} + x - 1} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 x^{3} - 1 = 0

解

4 x^{3} - 1 = 0 : x = \frac{1}{3 4}, x = - \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = - \frac{1}{2 3 4} - \frac{3}{2 3 4} i

x = \frac{1}{3 4}, x = - \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = - \frac{1}{2 3 4} - \frac{3}{2 3 4} i

验证解

找到无定义的点（奇点）:

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

将不在定义域的点与解相综合:

x = \frac{1}{3 4}, x = - \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = - \frac{1}{2 3 4} - \frac{3}{2 3 4} i

3 + (\frac{1}{x} - 2) = 30

0.1 = 1 - (1 - \frac{1}{x})^{25}

\frac{2}{x - 1} = \frac{4}{x + 2}

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

so l v e f or X, 4 Y = \frac{3 X}{X + 1}