$1.14 \frac{x^{4}+2 x^{2}-3 x+1}{x^{2}+\sqrt{3 x}+1}+2\left(\sqrt[6]{27 x^{3}}-\frac{1}{2}\right)$.
1.14 Let's denote the fraction as $A$, and the expression in parentheses as $B$, then the given expression will take the form $A+2 B$. Note that for $\sqrt{3 x}$ and $\sqrt[6]{27 x^{3}}$, only values $x \geqslant 0$ are permissible.

Using formula (1.9), we extract a complete square in the numerator of the fraction $A$:

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

Since $x \geqslant 0$, by equality (1.21) we have $3 x=(\sqrt{3 x})^{2}$. Then the obtained expression can be factored using formula (1.8) as a difference of squares:

$$
\left(x^{2}+1\right)^{2}-(\sqrt{3 x})^{2}=\left(x^{2}+1-\sqrt{3 x}\right)\left(x^{2}+1+\sqrt{3 x}\right)
$$

Therefore,

$$
A=\frac{\left(x^{2}+1-\sqrt{3 x}\right)\left(x^{2}+1+\sqrt{3 x}\right)}{x^{2}+\sqrt{3 x}+1}=x^{2}-\sqrt{3 x}+1
$$

Next, based on formula (1.20) we have $\sqrt[6]{27 x^{3}}=$ $=\sqrt[6]{(3 x)^{3}}=\sqrt{3 x}$, hence $B=\sqrt{3 x}-\frac{1}{2}$.

Thus,

$$
A+2 B=x^{2}-\sqrt{3 x}+1+2 \sqrt{3 x}-1=x^{2}+\sqrt{3 x}
$$

Answer: $x^{2}+\sqrt{3 x}$.