Question

\(P=\left(\dfrac{\sqrt{x}+1}{x-1}+\dfrac{x}{x-\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\) với x > 0; x \(\ne\) 1

a) Rút gọn P

b) Tìm GTNN của biểu thức M = P.\(\sqrt{x}\) khi x > 1

Answer 1

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

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

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

b. Áp dụng BĐT AM-GM

\(M=P\sqrt{x}=\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)

\(=(\sqrt{x}-1)+\frac{1}{\sqrt{x}-1}+2\geq 2\sqrt{(\sqrt{x}-1).\frac{1}{\sqrt{x}-1}}+2=2+2=4\)

Vậy $M_{\min}=4$ khi $\sqrt{x}-1=\frac{1}{\sqrt{x}-1}$

$\Rightarrow \sqrt{x}-1=0$

$\Leftrightarrow x=1$