\(A=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}\)
Có: \(3x^2+9x+7=3\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{4}=3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)
Vì: \(3\left(x+\frac{3}{2}\right)^2\ge0,\forall x\)
=> \(3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
=>\(\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}}\le40\)
=> \(1+\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{41}{4}}\le41\)
Vậy GTLN của A là \(\frac{81}{41}\) khi \(x=-\frac{3}{2}\)
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