a. \(y=\sqrt{x^2-6x+10}=\sqrt{x^2-6x+9+1}=\sqrt{\left(x-3\right)^2+1}\ge\sqrt{0+1}=1\)
\(\Rightarrow Min_y=1\Leftrightarrow x=3\)
b. \(y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+1}=\sqrt{\left(\dfrac{x}{3}\right)^2-2.\dfrac{x}{3}.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{24}{25}}=\sqrt{\left(\dfrac{x}{3}-\dfrac{1}{5}\right)^2+\dfrac{24}{25}}\ge\sqrt{0+\dfrac{24}{25}}=\sqrt{\dfrac{24}{25}}\)
\(\Rightarrow Min_y=\sqrt{\dfrac{24}{25}}\Leftrightarrow x=\dfrac{3}{5}\)
Giải:
a) \(y=\sqrt{x^2-6x+10}\)
\(\Leftrightarrow y=\sqrt{x^2-6x+9+1}\)
\(\Leftrightarrow y=\sqrt{\left(x^2-6x+9\right)+1}\)
\(\Leftrightarrow y=\sqrt{\left(x-3\right)^2+1}\ge1\)
\(\Leftrightarrow y_{Min}=1\)
\("="\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy ...
b) \(y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+1}\)
\(\Leftrightarrow y=\sqrt{\dfrac{x^2}{9}-\dfrac{2x}{15}+\dfrac{1}{25}+\dfrac{24}{25}}\)
\(\Leftrightarrow y=\sqrt{\left(\dfrac{x^2}{9}-\dfrac{2x}{15}+\dfrac{1}{25}\right)+\dfrac{24}{25}}\)
\(\Leftrightarrow y=\sqrt{\left(\dfrac{x}{3}-\dfrac{1}{5}\right)^2+\dfrac{24}{25}}\ge\dfrac{24}{25}\)
\(\Leftrightarrow y_{Min}=\dfrac{24}{25}\)
\("="\Leftrightarrow\dfrac{x}{3}-\dfrac{1}{5}=0\Leftrightarrow x=\dfrac{3}{5}\)
Vậy ...
