Question

Bài 1: \( P = x - 2 \) ( \( \min x \in \mathbb{Z} \) để \( P \in \mathbb{Z} \) )

\( \sqrt{x} - 1 \) ( Với \( x = 0, 2 \pm 1 \) )

Bài 2: \( Q = 2\sqrt{x} \) ( Tìm số tự nhiên \( x \) )

\( \sqrt{x} - 3 \) ( sao cho \( Q \) min )

Answer 1

Bài 1:

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

P nguyên khi 1 ⋮ `(\sqrt{x}-1)'

`=>\sqrt{x}-1∈Ư(1)={1;-1}`

`=>\sqrt{x}∈{2;0}`

`=>x∈{4;0}`

Bài 2: 

\(Q=\dfrac{2\sqrt{x}}{\sqrt{x}-3}=\dfrac{2\left(\sqrt{x}-3\right)+6}{\sqrt{x}-3}=2+\dfrac{6}{\sqrt{x}-3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-3\ge-3\)

\(\Rightarrow\dfrac{6}{\sqrt{x}-3}\ge\dfrac{6}{-3}=-2\Rightarrow Q\ge2+\left(-2\right)=0\)

Vậy `Q_(min)=0` khi `x=0`