Question

Tim GTLN cua bieu thuc: Q = \(\frac{-\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\)

Answer 1

ta có :  (\(\sqrt{x}\)-   2   )\(^2\)\(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4  \(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4 +   8\(\sqrt{x}\) \(\ge\)8\(\sqrt{x}\)

   \(\Leftrightarrow\)(\(\sqrt{x}\)+    2  )\(^2\)\(\ge\)8\(\sqrt{x}\)

\(\Leftrightarrow\)-(\(\sqrt{x}\)+    2  )\(^2\)\(\le\)-8\(\sqrt{x}\)

\(\Leftrightarrow\)Q  \(\le\)\(\frac{-8\sqrt{x}}{\sqrt{x}}\)=   (   -  8  )

Dấu ''   =   ''   xaye ra tại   x =  4