Đáp án: `P_{max}=\frac{2020}{2019}⇔x=2020`
Giải thích các bước giải:
Đặt `P=\frac{x^2}{x^2-2x+2020}(x∈R)`
-Nếu $x=0⇒P=0$
-Nếu $x\neq0$
`⇒\frac{1}{P}=\frac{x^2-2x+2020}{x^2}`
`=1-\frac{2}{x}+\frac{2020}{x^2}`
`=\frac{2019}{2020}+2020(\frac{1}{x^2}-2.\frac{1}{x}.\frac{1}{2020}+\frac{1}{2020^2})`
`=\frac{2019}{2020}+2020(\frac{1}{x}-\frac{1}{2020})^2`
Do `(\frac{1}{x}-\frac{1}{2020})^2≥0`
`⇒2020(\frac{1}{x}-\frac{1}{2020})^2≥0`
`⇒\frac{1}{P}=\frac{2019}{2020}+2020(\frac{1}{x}-\frac{1}{2020})^2≥\frac{2019}{2020}`
`⇒P≤\frac{2020}{2019}`
Dấu bằng xảy ra `⇔(\frac{1}{x}-\frac{1}{2020})^2=0`
`⇔\frac{1}{x}-\frac{1}{2020}=0⇔\frac{1}{x}=\frac{1}{2020}⇔x=2020(tm)`
So sánh cả $2$ trường hợp, ta được:
`P_{max}=\frac{2020}{2019}⇔x=2020`
