已知 $a_{1}>a_{2}>a_{3}>0$, 则使得 $\left(1-a_{i} x\right)^{2}<1(i=1,2,3)$ 都成立的 $x$ 的取值范围是 (
由 $\left(1-a_{i} x\right)^{2}<1$,

得 $1-2 a_{i} x+\left(a_{i} x\right)^{2}<1$,

即 $a_{i} \cdot x\left(a_{i} x-2\right)<0$.

又 $a_{1}>a_{2}>a_{3}>0$.

$\therefore 0<x<\frac{2}{a_{i}}$,

即 $x<\frac{2}{a_{1}}, x<\frac{2}{a_{2}}$ 且 $x<\frac{2}{a_{3}}$.

$\because \frac{2}{a_{3}}>\frac{2}{a_{2}}>\frac{2}{a_{1}}>0$

$\therefore 0<x<\frac{2}{a_{1}}$.