Question

Tìm GTLN của

B = \(\frac{3}{x^2+2x+5}\)

Answer 1

Ta có :

\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)

\(\Rightarrow B=\frac{3}{\left(x+1\right)^2+4}\)

Vì \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+4\ge4\)

\(\Rightarrow\frac{3}{\left(x+1\right)^2+4}\le\frac{3}{4}\)

\(\Rightarrow B\le\frac{3}{4}\)

Dấu " = " xảy ra khi x = - 1

Vậy MAX_B= 3 / 4 khi x = - 1

Answer 2

=3

chuan 100%