Question

Tìm giá trị nhỏ nhất của các biểu thức

H = \(\dfrac{-2021}{4x^2+4x+3}\)

I = \(\dfrac{-2019}{5x^2-2x+10}\)

Answer 1

1) \(4x^2+4x+3=\left(2x+1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{2021}{4x^2+4x+3}\le\dfrac{2021}{2}\Rightarrow H=-\dfrac{2021}{4x^2+4x+3}\ge-\dfrac{2021}{2}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{1}{2}\)

2) \(5x^2-2x+10=5\left(x^2-\dfrac{2}{5}x+\dfrac{1}{25}\right)+\dfrac{49}{5}=5\left(x-\dfrac{1}{5}\right)^2+\dfrac{49}{5}\ge\dfrac{49}{5}\)

\(\Rightarrow I=\dfrac{-2019}{5x^2-2x+10}\ge-\dfrac{10095}{49}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{5}\)

Answer 2

ghi lộn môn hc ạ