Question

Tính P = \(\left(\frac{\sqrt{x}}{\sqrt{x-1}}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne1\) Sửa \(\sqrt{x-1}\) thành \(\sqrt{x}-1\)

\(P=\left(\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

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

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