Question

[H Phương Nguyên]

Bài 1)tìm Min hay Max

a) G=\(\dfrac{2}{x^2+8}\)

b) H=\(\dfrac{-3}{x^2-5x+1}\)

Bài 2) Tìm Min hay Max

a)D=\(\dfrac{2x^2-16x+41}{x^2-8x+22}\)

b)E=\(\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}\)

c)G=\(\dfrac{3x^2-12x+10}{x^2-4x+5}\)

Answer 1

[Nguyễn Việt Lâm]

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)