ĐKXĐ : \(x\ge0\) và \(x\ne1\)
1 ) \(x=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\right)\left(\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=-\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)
\(=4\)
Thay \(x=4\) vào B ta được :
\(B=\frac{4+2}{4+\sqrt{4}+1}=\frac{6}{7}\)
2 ) \(A=\frac{1}{\sqrt{x}-1}-\frac{x-\sqrt{x}+3}{x\sqrt{x}-1}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
3 ) \(P=\frac{2}{x+\sqrt{x}+1}:\left(1-\frac{x+2}{x+\sqrt{x}+1}\right)\)
\(=\frac{2}{x+\sqrt{x}+1}:\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(=\frac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2}{\sqrt{x}-1}\)
Để \(P\le1\Leftrightarrow\frac{2}{\sqrt{x}-1}\le1\Leftrightarrow2\le\sqrt{x}-1\Leftrightarrow\sqrt{x}\ge3\Leftrightarrow x\ge9\)
Vậy \(x\ge9\) thì \(P\le1\)
