Question

tìm GTNN \(M=\dfrac{x+12}{\sqrt{x}-2}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)

\(M=\dfrac{x+12}{\sqrt{x}-2}=\dfrac{x-4+16}{\sqrt{x}-2}\)

\(=\sqrt{x}+2+\dfrac{16}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{16}{\sqrt{x}-2}+4\)

=>\(M>=2\cdot\sqrt{\left(\sqrt{x}-2\right)\cdot\dfrac{16}{\sqrt{x}-2}}+4=2\cdot4+4=12\)

Dấu '=' xảy ra khi \(\left(\sqrt{x}-2\right)^2=16\)

=>\(\left[{}\begin{matrix}\sqrt{x}-2=-4\\\sqrt{x}-2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(loại\right)\\\sqrt{x}=6\end{matrix}\right.\)

=>x=36(nhận)