Đặt A = \(9x+\dfrac{4x+3}{2x+1}\) (x\(\ne\dfrac{-1}{2}\))
= \(\dfrac{18x^2+13x+3}{2x+1}\)
<=> 2A = \(\dfrac{36x^2+26x+6}{2x+1}\)
= \(\dfrac{9\left(2x+1\right)^2-5\left(2x+1\right)+2}{2x+1}\)
= \(9\left(2x+1\right)+\dfrac{2}{2x+1}-5\)
Áp dụng bdt Co-si, ta có:
\(9\left(2x+1\right)+\dfrac{2}{2x+1}\ge2\sqrt{9\left(2x+1\right).\dfrac{2}{2x+1}}=6\sqrt{2}\)
=> 2A \(\ge6\sqrt{2}-5\)
<=> A \(\ge3\sqrt{2}-\dfrac{5}{2}\)
Dấu "=" xảy ra <=> x = \(\dfrac{-3+\sqrt{2}}{6}\)
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