Question

[Akira Nishihiko]

Tìm GTLN của các biểu thức:

a.\(\frac{1}{x^2+2}\)

b.\(\frac{1}{2x^2+5}\)

c.\(\frac{8}{\left(x-1\right)^2+4}\)

d.\(\frac{2}{6-x}\)

Answer 1

[Jason]

a) ta có \(x^2\ge0\Leftrightarrow x^2+2\ge2.\)

\(\frac{1}{x^2+2}\le\frac{1}{2}\) vậy GTLN là \(\frac{1}{2}\)

b) ta có \(2x^2\ge0\Leftrightarrow2x^2+5\ge5\)

\(\frac{1}{2x^2+5}\le\frac{1}{5}\) vậy GTLN là \(\frac{1}{5}\)

c) ta có \(\left(x-1\right)^2\ge0\Leftrightarrow\left(x-1\right)^2+4\ge4\)

\(\frac{8}{\left(x-1\right)^2+4}\le\frac{8}{4}\) vậy GTLN là \(\frac{8}{4}=2\)