Question

Rút gọn biểu thức :

\(a,\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)

\(b,\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)

Answer 1

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

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