Question

Bài 1: Cho hai biểu thức \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\) và \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\) với \(x\ge0\); \(x\ne9\)
Với \(x\in N\)* Tìm giá trị nhỏ nhất của \(P=A:B\)

Answer 1

Ta có: P = A:B

P = \(\dfrac{\sqrt{x}-2}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)

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

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{-3}{\sqrt{x}+1}\)

P = 1 + \(\dfrac{-3}{\sqrt{x}+1}\)

Theo ĐK, x \(\in\) N^* \(\Rightarrow x\ge1\)\(\Rightarrow\sqrt{x}\ge1\Rightarrow\sqrt{x}+1\ge2\)\(\Rightarrow\dfrac{1}{\sqrt{x}+1}\le\dfrac{1}{2}\Rightarrow\dfrac{-3}{\sqrt{x}+1}\ge\dfrac{-3}{2}\)\(\Rightarrow1+\dfrac{-3}{\sqrt{x}+1}\ge1+\dfrac{-3}{2}=\dfrac{-1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)(TMĐK)

Vậy Min(P) = \(\dfrac{-1}{2}\Leftrightarrow x=1\)