Question

[Trịnh Linh]

Tìm GTNN của các biểu thức sau:

\(C=\frac{x+10}{\sqrt{x}+3}\)  , x >= 0  

\(D= {x - 3 \sqrt {x}+3 \over \sqrt{x}-1}\); x >1

Answer 1

[Edogawa Conan]

Ta có: C = \(\frac{x+10}{\sqrt{x}+3}=\frac{x-9+19}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+19}{\sqrt{x}+3}=\sqrt{x}-3+\frac{19}{\sqrt{x}+3}\)

C = \(\sqrt{x}+3+\frac{19}{\sqrt{x}+3}-6\ge2.\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{19}{\left(\sqrt{x}+3\right)}}-6\)(bđt cosi)

C \(\ge2\sqrt{19}-6\)

Dấu "=" xảy ra <=> \(\sqrt{x}+3=\frac{19}{\sqrt{x}+3}\) <=> \(\left(\sqrt{x}+3\right)^2=19\)

<=> \(\orbr{\begin{cases}\sqrt{x}+3=\sqrt{19}\\\sqrt{x}+3=-\sqrt{19}\left(vn\right)\end{cases}}\) <=> \(\sqrt{x}=\sqrt{19}-3\) <=> \(x=22-6\sqrt{19}\)

Vậy MinC = \(2\sqrt{19}-6\) <=> \(x=22-6\sqrt{19}\)