BĐT AM-GM để xem à
\(A=\dfrac{\left(x+16\right)\left(x+9\right)}{x}=\dfrac{x^2+25x+144}{x}=x+25+\dfrac{144}{x}\)
Áp dụng BĐT AM-GM cho 2 số không âm
\(x+\dfrac{144}{x}\ge2\sqrt{\dfrac{x.144}{x}}\)
\(x+\dfrac{144}{x}\ge24\)
\(x+\dfrac{144}{x}+25\ge49\)
\(A\ge49\)
\(Min_A=49\)
\(A=\dfrac{x^2+25x+\left(3.4\right)^2}{x}=\dfrac{x^2+\left[49x-24x\right]+\left(3.4\right)^2}{x}=\dfrac{x^2-24x+\left(3.4\right)^2+49x}{x}\)\(A=\dfrac{\left(x-12\right)^2}{x}+49\ge49\)