Đặt $t={{3}^{x}}$> 0

Ta có: ${{3}^{x}}+{{9.3}^{-x}}<10\Leftrightarrow t+\frac{9}{t}<10\Leftrightarrow {{t}^{2}}=10t+9<0\Leftrightarrow 1

$={{3}^{0}}<{{3}^{x}}<{{3}^{2}}\Leftrightarrow 0

Mà $x\in \mathbb{Z}\Rightarrow x=1$${{d}_{2}}\cap \left( \alpha  \right)=B\left( 5;-2;1 \right)$

                                     ${{d}_{1}},{{d}_{2}}\left\{ \begin{align}  & x=1+4t \\  & y=-2t \\  & z=t \\ \end{align} \right.$