Question

\(p=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

tìm x để P < 1

Answer 1

ĐK : \(x\ge0;x\ne1\)

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

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

= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

P < 1 \(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}< 0\)

Mà 1 > 0 \(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Vậy để P < 1 thì \(0\le x< 1\)