Question

Tim GTLN cua bieu thuc

\(J=\frac{2010}{4x+20\sqrt{x}+30}\)

Answer 1

\(J=\frac{2010}{4x+20\sqrt{x}+30}\)

\(=\frac{2010}{\left(2\sqrt{x}\right)^2+2.2\sqrt{x}.5+25+5}\)

\(=\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)

\(A_{max}\Leftrightarrow\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)lớn nhất

\(\Rightarrow\left(2\sqrt{x}+5\right)^2+5\)nhỏ  nhất

\(\Rightarrow\left(2\sqrt{x}+5\right)^2\)nhỏ nhất 

Mà \(2\sqrt{x}+5\ge5\Rightarrow2\sqrt{x}+5=5\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\)

Với x = 0 \(\Rightarrow J_{max}=\frac{2010}{4.0+20\sqrt{0}+30}=\frac{2010}{30}=67\)