Question

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

a. Rút gọn biểu thức P

b. So sánh P với 3

c. So sánh P với \(\sqrt{P}\)

Answer 1

dkxd: x≥0; x≠1

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

\(=1:\left[\dfrac{1}{\sqrt{x}^3-1}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

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

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

Answer 2

c. Để \(\sqrt{P}\) có nghĩa \(P\ge0\Rightarrow\sqrt{x}>0\Rightarrow x\ge1\)

Vì \(x\ge1\) nên \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}>1\)

Ta có: P>1 \(P-\sqrt{P}=\sqrt{P}\left(\sqrt{P}-1\right)\left(1\right)\)

Vì P > 1\(\Rightarrow\sqrt{P}>1\Rightarrow\sqrt{P}-1>0\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\sqrt{P}\left(\sqrt{P}-1\right)>0\Rightarrow P-\sqrt{P}>0\Rightarrow P>\sqrt{P}\)

Answer 3

b. Ta có:

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

Vậy P >3