Question

a) \(\sqrt{\dfrac{x^2-2x+1}{x+2\sqrt{x}+1}}\)

b)\(\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)

Answer 1

a) \(\sqrt{\dfrac{x^2-2x+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(x-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2.\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\)

= \(\sqrt{\left(\sqrt{x}-1\right)^2}=|\sqrt{x}-1|\)

b) \(\dfrac{x-1}{\sqrt{y-1}}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)

= \(\dfrac{x-1}{\sqrt{y}-1}.\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\dfrac{\sqrt{y}-1}{x-1}\)