Solve for x

Solve for x

$\frac{1}{4x}+\frac{1}{3x}+\frac{1}{2x}=x$

$\begin{array}{rcll} \frac{1}{4x}+\frac{1}{3x}+\frac{1}{2x} &=& x \qquad | \qquad \cdot x\\\\ x\cdot \left( \frac{1}{4x}+\frac{1}{3x}+\frac{1}{2x} \right) &=& x \cdot x\\\\ \frac{x}{4x}+\frac{x}{3x}+\frac{x}{2x} &=& x^2\\\\ \frac{1}{4 }+\frac{1}{3 }+\frac{1}{2 } &=& x^2 \qquad | \qquad \cdot 4\\\\ 4\cdot \left( \frac{1}{4 }+\frac{1}{3 }+\frac{1}{2 } \right) &=& x^2 \cdot 4\\\\ \frac{4}{4 }+\frac{4}{3 }+\frac{4}{2 } &=& 4\cdot x^2 \\\\ 1+\frac{4}{3 }+2 &=& 4\cdot x^2 \\\\ 3+\frac{4}{3 } &=& 4\cdot x^2 \qquad | \qquad \cdot 3\\\\ 3\cdot \left( 3+\frac{4}{3 } \right)&=& 4\cdot x^2 \cdot 3\\\\ 3\cdot 3+\frac{3\cdot 4}{3 }&=& 3\cdot 4\cdot x^2\\\\ 9+ 4 &=& 12\cdot x^2\\\\ 13 &=& 12\cdot x^2\\\\ 12\cdot x^2 &=& 13\qquad | \qquad :12\\\\ x^2 &=& \frac{13}{12} \qquad | \qquad \sqrt{} \\\\ x &=& \pm \sqrt{ \frac{13}{12} } \\\\ x &=& \pm \sqrt{ \frac{13}{3\cdot 4} } \\\\ x &=& \pm \frac12 \cdot \sqrt{ \frac{13}{3} } \\\\ x &=& \pm 1.04083299973 \\\\ \mathbf{x_1} &\mathbf{=}& \mathbf{1.04083299973} \\\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1.04083299973} \\\\ \end{array}$

Solve for x:
13/(12 x) = x

13/(12 x) = 13/(12 x):
13/(12 x) = x

Multiply both sides by 12 x:
13 = 12 x^2

13 = 12 x^2 is equivalent to 12 x^2 = 13:
12 x^2 = 13

Divide both sides by 12:
x^2 = 13/12

Take the square root of both sides:
Answer: |  x = sqrt(13/3)/2               or                 x = -sqrt(13/3)/2

You have 1/24x³.

1/(24x^3)=x = {x=((i)/24^((1/4))), x=-((1/24^((1/4)))), x=-(((i)/24^((1/4)))), x=(1/24^((1/4)))}