Question

cho bt

P=(\(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)+\(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}\)):\(\dfrac{x+1+2\sqrt{x}}{x-1}\)

a)Rút gọn P

b)Tìm x để P<0

Answer 1

\(a,P=\left(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}\right):\dfrac{x+1+2\sqrt{x}}{x-1}\left(dk:x>0,x\ne1\right)\)

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

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

\(=\left(\sqrt{x}+1\right).\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\sqrt{x}-1\)

\(b,P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

So với \(dk:x>0\) \(\Rightarrow S=\left\{x|0< x< 1\right\}\)

Answer 2

đk với x>0,x khác 1