Question

the maximum value of \(\frac{\left(x^2+15\right)}{x^2+3}\)

Answer 1

\(\frac{x^2+15}{x^2+3}\)

\(=\frac{x^2+3+12}{x^2+3}\)

\(=\frac{x^2+3}{x^2+3}+\frac{12}{x^2+3}\)

\(=1+\frac{12}{x^2+3}\)

\(x^2\ge0\)

\(x^2+3\ge3\)

\(\frac{12}{x^2+3}\le4\)

\(1+\frac{12}{x^2+3}\le5\)

ĐS: 5