Question

Решите пожалуйста, срочно нужно!!!!

Answer 1

$(\frac{\sqrt{b}}{b-9} -\frac{\sqrt{b}}{b-6\sqrt{b}+9})*\frac{(3-\sqrt{b})^2}{2\sqrt{b}}+\frac{3}{\sqrt{b}+3} =0\\\\\\\\1)\ \frac{\sqrt{b}}{b-9} -\frac{\sqrt{b}}{b-6\sqrt{b}+9}=\frac{\sqrt{b}}{b-9} -\frac{\sqrt{b}}{(\sqrt{b}-3)^2}=\\\\=\frac{\sqrt{b}(\sqrt{b}-3)^2-\sqrt{b}(b-9)}{(b-9)(\sqrt{b}-3)^2} =\frac{-6b+18\sqrt{b}}{(b-9)(\sqrt{b}-3)^2} \\\\\\2)\ \frac{-6b+18\sqrt{b}}{(b-9)(\sqrt{b}-3)^2}*\frac{(3-\sqrt{b})^2}{2\sqrt{b}}=-\frac{3}{\sqrt{b}+3} \\\\\\3)\ -\frac{3}{\sqrt{b}+3}+\frac{3}{\sqrt{b}+3}=0$