Question

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

a) Ruts gọn P

b) So sánh \(\dfrac{1}{P}\) với 3

Answer 1

điều kiện \(x\ge0;x\ne1\)

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

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

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

b) ta có : \(\dfrac{1}{P}-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\left(x\ne1\right)\)

vậy \(\dfrac{1}{P}-3>0\Leftrightarrow\dfrac{1}{P}>3\)

Answer 2

Mysterious Person giúp mk

Answer 3

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

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

b: \(\dfrac{1}{P}-3=\dfrac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

=>1/P>3