Question

P= \(1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\) với x ≥0 x≠1 
a) rút gọn
b) so sánh P với 3

Answer 1

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

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

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

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

=>P>3

Answer 2

Lời giải:

a. 

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

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

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

\(=1:\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

b.

\(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{(\sqrt{x}-1)^2}{\sqrt{x}}>0\) với mọi $x>0; x\neq 1$

$\Rightarrow P>3$