tham khảo
\(A=\frac{4x+1}{4x^2+2}=\frac{4x^2+2}{4x^2+2}-\frac{4x^2-4x+1}{4x^2+2}=1-\frac{\left(2x-1\right)^2}{4x^2+2}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\)
\(A=\frac{4x+1}{4x^2+2}=\frac{-\left(2x^2+1\right)}{4x^2+2}+\frac{2x^2+4x+2}{4x^2+2}=\frac{-1}{2}+\frac{2\left(x+1\right)^2}{4x^2+2}\ge\frac{-1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-1\)
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