次の不等式を解け．

１ $1<\displaystyle\frac{1}{x-2}$

２ $\sqrt{2-x}\leq x+1$

２ \[\begin{eqnarray} && \sqrt{2-x}\leq x+1\\
&\Longleftrightarrow & 2-x\leq (x+1)^2\;\;{\rm and}\;\; 2-x\geq 0 \;\;{\rm and}\;\; x+1\geq 0\\
&\Longleftrightarrow & x^2+3x-1 \geq 0 \;\;{\rm and}\;\; x\leq 2 \;\;{\rm and}\;\; x\geq -1 \\
&\Longleftrightarrow & \left\{x\leq \frac{-3-\sqrt{13}}{2}\;\;{\rm or}\;\;\frac{-3+\sqrt{13}}{2} \leq x\right\} \;\;{\rm and}\;\; -1\leq x \leq 2\\
&\Longleftrightarrow &\frac{-3+\sqrt{13}}{2}\leq x\leq 2 \end{eqnarray}\]