Question

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Answer 1

$\displaystyle \tt \bigg(\frac{\sqrt{b}}{b-9}-\frac{\sqrt{b}}{b-6\sqrt{b}+9}\bigg)\cdot\frac{(3-\sqrt{b})^{2}}{2\sqrt{b}}+\frac{3}{\sqrt{b}+3}=\\\\\\=\bigg(\frac{\sqrt{b}}{(\sqrt{b}-3)(\sqrt{b}+3)}-\frac{\sqrt{b}}{(\sqrt{b}-3)^{2}}\bigg)\cdot\frac{(3-\sqrt{b})^{2}}{2\sqrt{b}}+\frac{3}{\sqrt{b}+3}=\\\\\\=\frac{\sqrt{b}\cdot(\sqrt{b}-3)-\sqrt{b}\cdot(\sqrt{b}+3)}{(\sqrt{b}-3)^{2}(\sqrt{b}+3)}\cdot\frac{(3-\sqrt{b})^{2}}{2\sqrt{b}}+\frac{3}{\sqrt{b}+3}=$  

$\displaystyle \tt =\frac{\sqrt{b}\cdot(\sqrt{b}-3-\sqrt{b}-3)}{2\sqrt{b}\cdot(\sqrt{b}+3)}+\frac{3}{\sqrt{b}+3} =-\frac{3}{\sqrt{b}+3}+\frac{3}{\sqrt{b}+3}=0$