6.148. $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}\right)=4$.

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6.148. $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}\right)=4$.
## Solution.

Domain of definition: $x \neq 0$.

Rewrite the equation as $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}-2\right)=0$ and let $x-\frac{1}{x}=y$. Then we get $2 y^{2}+3 y=0$, from which $y_{1}=0, y_{2}=-\frac{3}{2}$. Therefore, $x-\frac{1}{x}=0$ or $x-\frac{1}{x}=-\frac{3}{2}$.

Solving these equations, we find $x_{1,2}= \pm 1, x_{3}=-2, x_{4}=0.5$.

Answer: $x_{1,2}= \pm 1, x_{3}=-2, x_{4}=0.5$.