Ta có: \(B=\frac{x^2+2x+3}{x^2+2}\)
nên \(2B=\frac{2x^2+4x+6}{x^2+2}=\frac{\left(x^2+4x+4\right)+\left(x^2+2\right)}{x^2+2}=\frac{\left(x+2\right)^2}{x^2+2}+1\ge1\) với mọi \(x\)
\(\Rightarrow\) \(B\ge\frac{1}{2}\)
Dấu \(''=''\) xảy ra \(\Leftrightarrow\) \(\left(x+2\right)^2=0\) \(\Leftrightarrow\) \(x+2=0\) \(\Leftrightarrow\) \(x=-2\)
Vậy, \(B_{min}=\frac{1}{2}\) \(\Leftrightarrow\) \(x=-2\)
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Ta có: \(B=\frac{x^2+2x+3}{x^2+2}=\frac{2\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\) với mọi \(x\)
Dấu \(''=''\) xảy ra \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\)
Vậy, \(B_{max}=2\) \(\Leftrightarrow\) \(x=1\)