b) lấy kết quả rút gọn của câu A ta được
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}< 1.=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)
\(P=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
đề bài cho x>=0 ta suy ra luôn
\(x+2>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)
vậy x <1 thì P < 1
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right).\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}.\frac{\left(x+1\right)}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)