1. The solution set of the inequality $9^{x}-8^{x}-6^{x} \geqslant 1$ is
$\qquad$
$$
\begin{array}{l}
\text { I, 1. }[2,+\infty) \text {. } \\
\text { From } 9^{x}-8^{x}-6^{x} \geqslant 1 \\
\Leftrightarrow f(x)=\left(\frac{8}{9}\right)^{x}+\left(\frac{6}{9}\right)^{x}+\left(\frac{1}{9}\right)^{x} \leqslant 1 .
\end{array}
$$

Then $f(x)$ is monotonically decreasing.
Thus $f(x) \leqslant f(2)$.
Therefore, $x \geqslant 2$.