tìm GTNN của B=16x^2+4x+1/2X với x>0 B=(x+4).(x+9)/x với x>0 B=x/3+3/x-2 với x>2 B=(x+100)^2/x với x>0

Đáp án:

$\begin{array}{l}
B = \dfrac{{16{x^2} + 4x + 1}}{{2x}} = 8x + 2 + \dfrac{1}{{2x}}\\
Theo\,\text{Cô} - si:\\
8x + \dfrac{1}{{2x}} \ge 2.\sqrt {8x.\dfrac{1}{{2x}}}  = 2.\sqrt 4  = 4\\
 \Rightarrow 8x + \dfrac{1}{{2x}} + 2 \ge 6\\
 \Rightarrow B \ge 6\\
 \Rightarrow GTNN:B = 6\\
Khi:8x = \dfrac{1}{{2x}} \Rightarrow {x^2} = \dfrac{1}{{16}} \Rightarrow x = \dfrac{1}{4}\\
b)\\
B = \dfrac{{\left( {x + 4} \right)\left( {x + 9} \right)}}{x} = \dfrac{{{x^2} + 13x + 36}}{x}\\
 = x + 13 + \dfrac{{36}}{x} \ge 13 + 2\sqrt {x.\dfrac{{36}}{x}}  = 13 + 12 = 25\\
 \Rightarrow B \ge 25\\
 \Rightarrow GTNN:B = 25\\
Khi:x = \dfrac{{36}}{x} \Rightarrow {x^2} = 36 \Rightarrow x = 6\\
c)\\
B = \dfrac{x}{3} + \dfrac{3}{{x - 2}} = \dfrac{{x - 2}}{3} + \dfrac{2}{3} + \dfrac{3}{{x - 2}}\\
Theo\,\text{Cô} - si:\\
\dfrac{{x - 2}}{3} + \dfrac{3}{{x - 2}} \ge 2.\sqrt {\dfrac{{x - 2}}{3}.\dfrac{3}{{x - 2}}}  = 2\\
 \Rightarrow \dfrac{{x - 2}}{3} + \dfrac{2}{3} + \dfrac{3}{{x - 2}} \ge 2 + \dfrac{2}{3} = \dfrac{8}{3}\\
 \Rightarrow B \ge \dfrac{8}{3}\\
 \Rightarrow GTNN:B = \dfrac{8}{3}\\
Khi:\dfrac{{x - 2}}{3} = \dfrac{3}{{x - 2}}\\
 \Rightarrow {\left( {x - 2} \right)^2} = 9\\
 \Rightarrow x - 2 = 3\\
 \Rightarrow x = 5\\
d)\\
B = \dfrac{{{{\left( {x + 100} \right)}^2}}}{x} = \dfrac{{{x^2} + 200x + 10000}}{x}\\
 = x + 200 + \dfrac{{10000}}{x} \ge 200 + 2\sqrt {10000}  = 400\\
 \Rightarrow GTNN:B = 400\\
Khi:{x^2} = 10000 \Rightarrow x = 100
\end{array}$