2.136. $\frac{1-b}{\sqrt{b}} \cdot x^{2}-2 x+\sqrt{b} ; \quad x=\frac{\sqrt{b}}{1-\sqrt{b}}$.

2.136. $\frac{1-b}{\sqrt{b}} \cdot x^{2}-2 x+\sqrt{b} ; \quad x=\frac{\sqrt{b}}{1-\sqrt{b}}$.
## Solution.

Domain of definition: $0<b \neq 1$.

$$
\begin{aligned}
& \frac{1-b}{\sqrt{b}} \cdot\left(\frac{\sqrt{b}}{1-\sqrt{b}}\right)^{2}-2 \cdot \frac{\sqrt{b}}{1-\sqrt{b}}+\sqrt{b}=\frac{(1-\sqrt{b})(1+\sqrt{b})}{\sqrt{b}} \cdot \frac{b}{(1-\sqrt{b})^{2}}-\frac{2 \sqrt{b}}{1-\sqrt{b}}+ \\
& +\sqrt{b}=\frac{(1+\sqrt{b}) \sqrt{b}}{1-\sqrt{b}}-\frac{2 \sqrt{b}}{1-\sqrt{b}}+\sqrt{b}=\frac{\sqrt{b}+b}{1-\sqrt{b}}-\frac{2 \sqrt{b}}{1-\sqrt{b}}+\sqrt{b}= \\
& =\frac{\sqrt{b}+b-2 \sqrt{b}+\sqrt{b}-b}{1-\sqrt{b}}=0 .
\end{aligned}
$$

Answer: 0.