Question

4 sobre3x + 2x = a 1 sobre 2

 

Answer 1

Temos que:

 

$\dfrac{4}{3\text{x}}+2\text{x}=\dfrac{1}{2}$

 

Como $\text{mmc}(3\text{x}, 1, 2)=3\text{x}\cdot2=6\text{x}$, segue que:

 

$\dfrac{4\cdot2+2\text{x}\cdot3\text{x}\cdot2}{6\text{x}}=\dfrac{1\cdot3\text{x}}{6\text{x}}$

 

$8+12\text{x}^2=6\text{x}$

 

$12\text{x}^2-6\text{x}+8=0$

 

$\text{x}=\dfrac{-(-6)\pm\sqrt{(-6)^2-4\cdot12\cdot8}}{2\cdot1}=\dfrac{6\pm\sqrt{-348}}{2}$

 

Logo, não há solução real.

Answer 2

$\frac{4}{3x} + 2x = \frac{1}{2} \\\\ \frac{4}{3x_{/2}} + \frac{2x}{1_{/6x}} = \frac{1}{2_{/3x}} \\\\ 8 + 12x^2 = 3x \\ 12x^2 - 3x + 8 = 0 \\ \Delta = 9 - 384 \\ \Delta = - 375$

 

 Como $\Delta < 0$, $x \notin \mathbb{R}$