\(A=\dfrac{\left(x+16\right)\left(x+9\right)}{x}\)
\(A=\dfrac{x^2+25x+144}{x}\)
Vì x>0 nên ta được quyền rút gọn
\(A=x+25+\dfrac{144}{x}\)
Vì x>0 nên \(\dfrac{144}{x}>0\)
Áp dụng BĐT AM-GM cho \(x+\dfrac{144}{x}\left(x>0\right)\), ta có:
\(\dfrac{x+\dfrac{144}{x}}{2}\ge\sqrt{\dfrac{x.144}{x}}\)
\(x+\dfrac{144}{x}\ge2.\sqrt{144}\)
\(x+\dfrac{144}{x}\ge24\)
\(A=x+\dfrac{144}{x}+25\ge24+25\)
Vậy MinA =49 khi \(x=\dfrac{144}{x}\)
\(x=\dfrac{144}{x}\)
\(x^2=144\)
\(x=\pm12\)
Chọn nghiệm x=12 ( x>0)
Vậy: MinA=49 khi x=12