với x≥0 và x≠1 cho biểu thức P=\(\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right)\)
a)rút gọn P
b)tính giá trị của x để P<\(\frac{1}{2}\)
c)tìm giá trị của x để P=\(\frac{1}{3}\)
d)tìm x nguyên để P nguyên
e)tìm giá trị nhỏ nhất của P
\(P=\left(\frac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\frac{1}{x+1}\right)\)
\(=\left(\frac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{\left(x+1\right)}\right)=\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(x+1\right)}{\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(P< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
\(P=\frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{3}\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}=4\Rightarrow x=4\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
\(P\in Z\Leftrightarrow\sqrt{x}+1=Ư\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1\right\}\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow x=0\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\Rightarrow P\sqrt{x}+P=\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}\left(P-1\right)=-P-1\Rightarrow\sqrt{x}=\frac{P+1}{1-P}\)
\(\sqrt{x}\ge0\Rightarrow\frac{P+1}{1-P}\ge0\Rightarrow-1\le P\le1\)
\(\Rightarrow P_{min}=-1\) khi \(x=0\)