a. ĐKXĐ: \(x\ge0\) ; \(x\ne1\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-x}\right):\left(\frac{1}{\sqrt{x}+1}-\frac{2}{1-x}\right)\)
= \(\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
=\(\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\sqrt{x}-1\)
= \(\frac{x+1}{\sqrt{x}}\)
b. ĐKXĐ: \(x\ge0\) ; \(x\ne1\)
Ta có: \(x=7-4\sqrt{3}\)
= \(\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\)(thỏa mãn ĐKXĐ) vào P ta được:
P= \(\frac{\left(2-\sqrt{3}\right)^2+1}{\sqrt{\left(2-\sqrt{3}\right)^2}}\) = \(\frac{\left(2-\sqrt{3}\right)^2+1}{\left|2-\sqrt{3}\right|}\)= \(\frac{7-4\sqrt{3}+1}{2-\sqrt{3}}\) = \(\frac{8-4\sqrt{3}}{2-\sqrt{3}}\)
= \(\frac{4\left(2-\sqrt{3}\right)}{2-\sqrt{3}}\)=4
