Question

Answer 1

a) 

\(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+2}{1-2\sqrt{x}}\\ =\left(\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\dfrac{1-2\sqrt{x}}{\sqrt{x}+2}\\ =\dfrac{x-4\sqrt{x}+4+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{1-2\sqrt{x}}{\sqrt{x}+2}\\ =\dfrac{x+4\sqrt{x}+4}{x-4}.\dfrac{1-2\sqrt{x}}{\sqrt{x}+2}\\ =\dfrac{\left(\sqrt{x}+2\right)^2.\left(1-2\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2.\left(\sqrt{x}-2\right)}\\ =\dfrac{1-2\sqrt{x}}{\sqrt{x}-2}\)

b) `A=-1/3`

`<=> (1-2\sqrtx)/(\sqrtx-2)=-1/3`

`<=> 3(1-2\sqrtx)=-1(\sqrtx-2)`

`<=>3-6\sqrtx=-\sqrtx+2`

`<=>-5\sqrtx=-1`

`<=>\sqrtx=1/5`

`<=>x=1/25`