\(A=\dfrac{3x^2+12x+17}{x^2+4x+5}=\dfrac{3\left(x^2+4x+5\right)+2}{x^2+4x+5}=3+\dfrac{2}{x^2+4x+5}\)
Ta có: \(x^2+4x+5=x^2+4x+4+1=\left(x+2\right)^2+1\ge1\)
\(\Rightarrow\dfrac{2}{x^2+4x+5}\le2\Rightarrow A\le3+2=5\)
\(\Rightarrow A_{max}=5\) khi \(x=-2\)
đề thế này á?
\(A=3x^2+12x+\dfrac{17}{x^2}+4x+5\)
tìm GTNN
\(\sqrt{x+2\left(1+\sqrt{x+1}\right)}+\sqrt{x+2\left(1-\sqrt{x+1}\right)}\)