Question

Tìm min các biểu thức sau và giá trị x A=\(\dfrac{x^2+2x+17}{2\left(x+1\right)}\)

vs x>-1

giúp mk vs mk cần gâp lắm thank trc nha :)

Answer 1

\(A=\dfrac{x^2+2x+17}{2\left(x+1\right)}=\dfrac{x^2-6x+9+8x+8}{2\left(x+1\right)}\\ =\dfrac{\left(x-3\right)^2+8\left(x+1\right)}{2\left(x+1\right)}=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\)

Vì \(\left(x-3\right)^2\ge0;2\left(x+1\right)>0\) (do \(x>-1\))

\(\Rightarrow\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}\ge0\Leftrightarrow A=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\ge4\)

Dấu "=" xảy ra khi \(x=3\)

Vậy....