a.
\(y=\sqrt{x^2-2x+1+4}=\sqrt{\left(x-1\right)^2+4}\ge\sqrt{4}=2\)
\(y_{min}=2\) khi \(x=1\)
b.
\(y=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}=\sqrt{\dfrac{1}{36}\left(9x^2-6x+1\right)+\dfrac{35}{36}}=\sqrt{\dfrac{1}{36}\left(3x-1\right)^2+\dfrac{35}{36}}\ge\dfrac{\sqrt{35}}{6}\)
\(y_{min}=\dfrac{\sqrt{35}}{6}\) khi \(x=\dfrac{1}{3}\)
a)Ta có \(y=\sqrt{x^2-2x+5}=\sqrt{x^2-2x+1+4}=\sqrt{\left(x-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = xảy ra <=> x = 1.
b)Ta có \(y=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+\dfrac{1}{36}+\dfrac{35}{36}}=\sqrt{\left(\dfrac{x}{2}-\dfrac{1}{6}\right)^2+\dfrac{35}{36}}\ge\sqrt{\dfrac{35}{36}}\)
Dấu = xảy ra <=> \(x=\dfrac{1}{3}\)
