Question

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

Tìm x thuộc N để P đạt min

Answer 1

ĐK: x \(\in\) N, x \(\ne\) 0 \(\Rightarrow\) x \(\ge\) 1

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

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

P = 1 + \(\dfrac{1}{\sqrt{x}}\)

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

Vì theo ĐK, x\(\ge\)1\(\Rightarrow\sqrt{x}\ge1\)\(\Rightarrow-\dfrac{1}{\sqrt{x}}\ge-1\)\(\Rightarrow1-\left(-\dfrac{1}{\sqrt{x}}\right)\ge1-\left(-1\right)=2\)

Dấu "=" xảy ra \(\Leftrightarrow-\dfrac{1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)(T/m ĐK)

Vậy Min(P) = 2\(\Leftrightarrow x=1\)