Question

Tìm GTLN của:

A=1/x^2+2

B=-|x+2015|+4

Answer 1

\(A=\dfrac{1}{x^2+2}\)

Ta có \(x^2+2\ge2\Leftrightarrow\dfrac{1}{x^2+2}\le\dfrac{1}{2}\)

Vậy \(A_{max}=\dfrac{1}{2}\Leftrightarrow x=0\)

\(B=-\left|x+2015\right|+4\le4\\ B_{max}=4\Leftrightarrow x+2015=0\Leftrightarrow x=-2015\)

Answer 2

x2+2≥2⇔1x2+2≤12x2+2≥2⇔1x2+2≤12

Vậy