Question

Cho P = \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\) với x > 0

a, Rút gọn P

b, Tìm x sao cho P < 0

Answer 1

đkxđ: x > 0\(a,P=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, với x > 0 thì P < 0 \(\Leftrightarrow\) \(\frac{\sqrt{x}-1}{\sqrt{x}}< 0\) Do\(\sqrt{x}>0\) \(\Rightarrow\) \(\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\) . Kết hợp với đkxđ \(\Rightarrow\) 0 < x < 1 thì P < 0