Question

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

a)rút gọn

b)tìm x để P<0

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

\(P< 0\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Rightarrow\sqrt{x}-1< 0\Rightarrow0< x< 1\)