CM: 2020$x{2}$-4x+4 không âm với mọi x€R 25/11/2021 Bởi Peyton CM: 2020$x{2}$-4x+4 không âm với mọi x€R
\(2020x^2-4x+4)\) \(=x^2-4x+4+2019x^2\) \(=(x^2-4x+4)+2019x^2\) \(=(x-2)^2+2019x^2\) \(\Rightarrow\begin{cases}(x-2)^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\\x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\end{cases}\) \(\Rightarrow\begin{cases}(x-2)^2\geq 0\,\,\,\,\forall\,\,\,x\in\mathbb{R}\\2019x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\end{cases}\) \(\Rightarrow (x-2)^2+2019x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\) Vậy giá trị của biểu thức không âm \(x\in\mathbb{R}\) Bình luận
$\quad 2020x^2 – 4x + 4$ $= (x^2 – 4x + 4) + 2019x^2$ $= (x-2)^2 + 2019x^2$ Ta có: $\quad \begin{cases}(x-2)^2 \geq 0\quad \forall x\\x^2 \geq 0\quad \forall x\end{cases}$ Do đó: $(x-2)^2 + 2019x^2 \geq 0\forall x$ Vậy $2020x^2 – 4x + 4$ không âm $\forall x\in\Bbb R$ _____________________________________________________________ $\quad 2020x^2 – 4x + 4$ $= 2020\left(x^2 – \dfrac{1}{505}x + \dfrac{1}{505}\right)$ $= 2020\left(x^2 – 2\cdot\dfrac{1}{1010}x + \dfrac{1}{1010^2}\right) + \dfrac{2019}{505}$ $= 2020\left(x – \dfrac{1}{1010}\right)^2 + \dfrac{2019}{505}$ Ta có: $\quad \left(x – \dfrac{1}{1010}\right)^2\geq 0\quad \forall x$ $\to 2020\left(x – \dfrac{1}{1010}\right)^2 + \dfrac{2019}{505} \geq \dfrac{2019}{505} >0$ Hay $2020x^2 – 4x + 4 >0$ Vậy $2020x^2 – 4x + 4$ luôn dương $\forall x \in\Bbb R$ Bình luận
\(2020x^2-4x+4)\)
\(=x^2-4x+4+2019x^2\)
\(=(x^2-4x+4)+2019x^2\)
\(=(x-2)^2+2019x^2\)
\(\Rightarrow\begin{cases}(x-2)^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\\x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\end{cases}\)
\(\Rightarrow\begin{cases}(x-2)^2\geq 0\,\,\,\,\forall\,\,\,x\in\mathbb{R}\\2019x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\end{cases}\)
\(\Rightarrow (x-2)^2+2019x^2\geq 0\,\,\,\forall\,\,\,x\in\mathbb{R}\)
Vậy giá trị của biểu thức không âm \(x\in\mathbb{R}\)
$\quad 2020x^2 – 4x + 4$
$= (x^2 – 4x + 4) + 2019x^2$
$= (x-2)^2 + 2019x^2$
Ta có:
$\quad \begin{cases}(x-2)^2 \geq 0\quad \forall x\\x^2 \geq 0\quad \forall x\end{cases}$
Do đó:
$(x-2)^2 + 2019x^2 \geq 0\forall x$
Vậy $2020x^2 – 4x + 4$ không âm $\forall x\in\Bbb R$
_____________________________________________________________
$\quad 2020x^2 – 4x + 4$
$= 2020\left(x^2 – \dfrac{1}{505}x + \dfrac{1}{505}\right)$
$= 2020\left(x^2 – 2\cdot\dfrac{1}{1010}x + \dfrac{1}{1010^2}\right) + \dfrac{2019}{505}$
$= 2020\left(x – \dfrac{1}{1010}\right)^2 + \dfrac{2019}{505}$
Ta có:
$\quad \left(x – \dfrac{1}{1010}\right)^2\geq 0\quad \forall x$
$\to 2020\left(x – \dfrac{1}{1010}\right)^2 + \dfrac{2019}{505} \geq \dfrac{2019}{505} >0$
Hay $2020x^2 – 4x + 4 >0$
Vậy $2020x^2 – 4x + 4$ luôn dương $\forall x \in\Bbb R$