Dạng 5: Bài toán nâng cao có đáp án

Để $\frac{3}{u-2}$nguyên thì $3⋮u-2\Rightarrow u-2\in Ư\left(3\right)=\{\pm 1;\pm 3\}$

$\frac{3u^2-2u+1}{3u+1} nguyên thì 2⋮3u+1\Rightarrow 3u+1\in Ư \left(2\right)=\{\pm 1;\pm 2\}$

$A=\frac{2019x^2-2.x.2019+{2019}^2}{2019x^2}=\frac{{(x-2019)}^2}{2019x^2}+\frac{2018x^2}{2019x^2}=\frac{{(x-2019)}^2}{2019x^2}+\frac{2018}{2019}$

$$\begin{gathered} Vì \frac{{(x-2019)}^2}{2019x^2}⩾0\forall x>0\Rightarrow \frac{{(x-2019)}^2}{2019x^2}+\frac{2018}{2019}⩾\frac{2018}{2019} \\ Dấu "=" xảy ra khi x-2019=0\Rightarrow x=2019 \\ Vậy Min A=\frac{2018}{2019} khi x=2019 \end{gathered}$$

$B=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}=1+\frac{10}{3{(x+{\displaystyle \frac{3}{2}})}^2+\frac{1}{4}}$

$Để B lớn nhất \Rightarrow 3.{(x+\frac{3}{2})}^2+\frac{1}{4} nhỏ nhất$

$$\begin{gathered} Mà 3.{(x+\frac{3}{2})}^2+\frac{1}{4}\ge \frac{1}{4} vì 3.{(x+\frac{3}{2})}^2\ge 0\forall x\in R \\ Dấu "=" xảy ra khi x+\frac{3}{2}=0\Rightarrow x=\frac{-3}{2} \\ Vậy Max B=41 khi x=\frac{-3}{2} \end{gathered}$$