Question

Rút gọn:

\(A=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\frac{1}{\sqrt{x-1}}\)

Tính giá trị của x tại \(x=19-8\sqrt{3}\)

Answer 1

ĐKXĐ: \(x>1\)

\(A=\left(\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}}+\frac{\sqrt{\left(\sqrt{x}+1\right)^2}}{\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}}\right)\sqrt{x-1}\)

\(=\left(\frac{\sqrt{x}-1}{\sqrt{x-1}}+\frac{\sqrt{x}+1}{\sqrt{x-1}}\right).\sqrt{x-1}\)

\(=\frac{2\sqrt{x}.\sqrt{x-1}}{\sqrt{x-1}}=2\sqrt{x}\)

\(A=2\sqrt{19-8\sqrt{3}}=2\sqrt{\left(4-\sqrt{3}\right)^2}=2\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)