Question

tìm GTLN của :A=\(\frac{10}{x^2-4x+6}\)

Answer 1

\(A=\frac{10}{\left(x-2\right)^2+2}\)

Do \(\left(x-2\right)^2+2\ge2>0\Rightarrow\frac{10}{\left(x-2\right)^2+2}\le\frac{10}{2}=5\)

\(A_{max}=5\) khi \(x=2\)