Question

Answer 1

`1)`\(A=\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(A=\left(\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(A=\left(\dfrac{x+2\sqrt{x}-7-x-3\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(A=\dfrac{-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)

`2)`\(A=-2\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=-2\)

\(\Leftrightarrow\sqrt{x}-1=-2\sqrt{x}+6\)

\(\Leftrightarrow3\sqrt{x}=7\)

\(\Leftrightarrow x=\dfrac{49}{9}\) ( tm )