Question

Answer 1

1) \(< =>A=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(< =>A=\sqrt{x}-1\)
2) Ta có \(x=8-2\sqrt{7}< =>x=7-2\sqrt{7}+1< =>x=\left(\sqrt{7}-1\right)^2\)
\(=>A=\sqrt{x}-1< =>A=\sqrt{\left(\sqrt{7}-1\right)^2}-1\)
\(< =>A=\left|\sqrt{7}-1\right|-1< =>A=\sqrt{7}-1-1< =>A=\sqrt{7}-2\)