Question

Tim GTNN va GTLN

b) \(B=\dfrac{2x+4}{x^2+2}\)

c) \(C=\dfrac{4x^2-4x-17}{\left(x-2\right)^2}\)

Answer 1

\(B=\dfrac{2x+4}{x^2+2}\)

\(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge2\)

\(\Rightarrow\dfrac{2x+4}{x^2+2}\le\dfrac{2x+4}{2}\)

Dấu "=" xảy ra khi:

\(x^2=0\Rightarrow x=0\)

\(\Rightarrow MAX_B=\dfrac{2.0+4}{0^2+2}=\dfrac{4}{2}=2\)

\(C=\dfrac{4x^2-4x-7}{\left(x-2\right)^2}\)

\(\left(x-2\right)^2\ne0\)

\(\left(x-2\right)^2\ge0\)

\(C=\dfrac{4x^2-4x-7}{\left(x-2\right)^2}\le\dfrac{4x^2-4x-7}{1}\)

\(MAX_C=\dfrac{4.3^2-4.3-7}{\left(3-2\right)^2}=\dfrac{17}{1}=17\)